Start in science. Solving equations in integers
Municipal educational institution
Savrushskaya average secondary school
Pokhvistnevsky district, Samara region
Abstract in mathematics on the topic:
"Equations with two
unknown
in whole numbers"
Completed by: Kolesova Tatyana
Staroverova Nina
at 10th grade students
Municipal educational institution Savrushskaya secondary school
Pokhvistnevsky district
Samara region.
Supervisor: Yatmankina Galina Mikhailovna
mathematics teacher.
Savrukha 2011
Introduction.________________________________________________3
1. Historical background _______________________________________5
1.1 Theorems on the number of solutions of linear Diophantine equations___6
1.2 Algorithm for solving equations in integers_________________ 6
1.3 Methods for solving equations______________________________ 7
Chapter 2. Application of methods for solving equations.
1. Problem solving_____________________________________________ 8
2.1 Solving problems using the Euclidean algorithm________________ 8
2.2 Method of enumerating options________________________________ 9
2.3 Factorization method___________________________ 9
2.4 Residual method________________________________________________ 12
2. Exam level tasks___________________________ 13
Conclusion________________________________________________ 16
List of references ________________________________________ 17
"Who controls the numbers,
He rules the world"
Pythagoras.
Introduction.
Situation analysis: Diophantine equations are a topical topic in our time, since the solution of equations, inequalities, and problems that reduce to solving equations in integers using estimates for variables is found in various mathematical collections and collections of the Unified State Examination.
Having studied different ways solutions quadratic equation with one variable in class, we were interested in understanding how equations with two variables are solved. Such tasks are found at Olympiads and in Unified State Examination materials.
In this academic year Eleventh graders will have to take the Unified State Exam in Mathematics, where the KIMs are compiled according to a new structure. There is no part “A”, but tasks have been added to part “B” and part “C”. The compilers explain the addition of C6 by the fact that in order to enter a technical university you need to be able to solve tasks of such high level complexity.
Problem: Deciding approximate options Unified State Exam assignments, we noticed that the most common tasks in C6 are solving equations of the first and second degrees in integers. But we do not know how to solve such equations. In this regard, it became necessary to study the theory of such equations and the algorithm for solving them.
Target: Master the method of solving equations with two unknowns of the first and second degree in integers.
Tasks: 1) Study educational and reference literature;
2) Collect theoretical material on methods for solving equations;
3) Analyze the algorithm for solving equations of this type;
4) Describe the solution.
5) Consider a number of examples using this technique.
6) Solve equations with two variables in integers from
materials of the Unified State Exam-2010 C6.
Object of study : Solving equations
Subject of research : Equations with two variables in integers.
Hypothesis: This topic is of great practical importance. In the school mathematics course, equations with one variable and various ways their decisions. Needs educational process require students to know and be able to solve simple equations with two variables. Therefore, increased attention to this topic is not only justified, but is also relevant in the school mathematics course.
This work can be used to study this topic in elective classes for students, in preparation for final and entrance exams. We hope that our material will help high school students learn to solve equations of this type.
Chapter 1. Theory of equations with two variables in integers.
1. Historical background.
Diophantus and the history of Diophantine equations .
Solving equations in integers is one of the oldest mathematical problems. This area of mathematics reached its greatest flourishing in Ancient Greece. The main source that has come down to our time is the work of Diophantus - “Arithmetic”. Diophantus summarized and expanded the experience accumulated before him in solving indefinite equations in integers.
History has preserved for us few features of the biography of the remarkable Alexandrian algebraist Diophantus. According to some sources, Diophantus lived until 364 AD. Only the unique biography of Diophantus is known for certain, which, according to legend, was carved on his tombstone and presented a puzzle task:
“God sent him to be a boy for a sixth of his life; adding to this the twelfth part, He covered his cheeks with down; after the seventh part, He lit the light of marriage for him and five years after marriage gave him a son. Alas! An unfortunate late child, having reached the measure of half of his father's full life, he was carried away by a merciless fate. Four years later, consoling the grief that befell him with the science of numbers, he [Diophantus] ended his life” (approximately 84 years old).
This puzzle serves as an example of the problems that Diophantus solved. He specialized in solving problems in integers. Such problems are currently known as Diophantine problems.
The most famous, solved by Diophantus, is the problem of “decomposition into two squares.” Its equivalent is the well-known Pythagorean theorem. This theorem was known in Babylonia, perhaps it was also known in Ancient Egypt, but it was first proven in the Pythagorean school. This was the name of a group of philosophers interested in mathematics named after the founder of the school of Pythagoras (c. 580-500 BC)
The life and work of Diophantus took place in Alexandria, he collected and solved known problems and came up with new ones. He later combined them in a great work called Arithmetic. Of the thirteen books that made up Arithmetic, only six survived into the Middle Ages and became a source of inspiration for Renaissance mathematicians.
1.1 Theorems on the number of solutions to a linear Diophantine equation.
We present here the formulations of theorems on the basis of which an algorithm for solving indeterminate first-degree equations of two variables in integers can be compiled.
Theorem 1. If in an equation , , then the equation has at least one solution.
Theorem 2. If in the equation , and With is not divisible by , then the equation has no integer solutions.
Theorem 3. If in the equation , and , then it is equivalent to the equation in which .
Theorem 4. If in the equation , , then all integer solutions to this equation are contained in the formulas:
Where x 0, y 0
1.2. Algorithm for solving equations in integers.
The theorems formulated allow us to compose the following algorithm solutions in integers to equations of the form .
1. Find the largest common divisor numbers a And b ,
if and With is not divisible by , then the equation has no integer solutions;
if and , then
2. Divide the equation term by term, obtaining an equation in which .
3. Find the whole solution ( x 0, y 0) equations by representing 1 as a linear combination of numbers and ;
4. Compose general formula integer solutions to this equation
Where x 0, y 0– an integer solution to the equation, - any integer.
1.3 Methods for solving equations
When solving equations in integers and natural numbers The following methods can be roughly distinguished:
1. Method of enumerating options.
2. Euclidean algorithm.
3. Continued fractions.
4. Factorization method.
5. Solving equations in integers as squares with respect to some variable.
6. Residue method.
7. Infinite descent method.
Chapter 2. Application of methods for solving equations
1. Examples of solving equations.
2.1 Euclidean algorithm.
Problem 1 . Solve equation in integers 407 X – 2816y = 33.
Let's use the compiled algorithm.
1. Using the Euclidean algorithm, we find the greatest common divisor of the numbers 407 and 2816:
2816 = 407 6 + 374;
407 = 374 1 + 33;
374 = 33 11 + 11;
Therefore (407.2816) = 11, with 33 divisible by 11
2. Divide both sides of the original equation by 11, we get equation 37 X – 256y= 3, with (37, 256) = 1
3. Using the Euclidean algorithm, we find a linear representation of the number 1 through the numbers 37 and 256.
256 = 37 6 + 34;
Let us express 1 from the last equality, then successively ascending the equalities we will express 3; 34 and substitute the resulting expressions into the expression for 1.
1 = 34 – 3 11 = 34 – (37 – 34 1) 11 = 34 12 – 37 11 = (256 – 37 6) 12 – 37 11 =
– 83·37 – 256·(–12)
Thus, 37·(–83) – 256·(–12) = 1, therefore a pair of numbers x 0= – 83 and y 0= – 12 is the solution to equation 37 X – 256y = 3.
4. Let us write down the general formula for solutions to the original equation
Where t- any integer.
2.2 Method of enumerating options.
Task 2. Rabbits and pheasants sit in a cage; they have 18 legs in total. Find out how many of both are in the cell?
Solution: An equation is drawn up with two unknown variables, in which x is the number of rabbits, y is the number of pheasants:
4x + 2y = 18, or 2x + y = 9.
Let's express at through X : y = 9 – 2x.
Thus, the problem has four solutions.
Answer: (1; 7), (2; 5), (3; 3), (4; 1).
2.3 Factorization method.
Enumerating options when finding natural solutions to an equation with two variables turns out to be very labor-intensive. Moreover, if the equation has whole solutions, then it is impossible to enumerate them, since there are an infinite number of such solutions. Therefore, we will show one more technique - factorization method.
Task 3. Solve equation in whole numbers y 3 - x 3 = 91.
Solution. 1) Using abbreviated multiplication formulas, we factorize the right side of the equation:
(y - x)(y 2 + xy + x 2) = 91……………………….(1)
2) Let’s write down all the divisors of the number 91: ± 1; ± 7; ± 13; ± 91
3) Conduct research. Note that for any integers x And y number
y 2 + yx + x 2 ≥ y 2 - 2|y ||x | + x 2 = (|y | - |x |) 2 ≥ 0,
therefore, both factors on the left side of the equation must be positive. Then equation (1) is equivalent to a set of systems of equations:
; 
; 
; 
4) Having solved the systems, we obtain: the first system has solutions (5; 6), (-6; -5); third (-3; 4),(-4; 3); the second and fourth have no solutions in integers.
Answer: equation (1) has four solutions (5; 6); (-6; -5); (-3; 4); (-4;3).
Task 4. Find all pairs of natural numbers satisfying the equation
Solution. Let's factorize the left side of the equation and write the equation in the form
.
Because The divisors of the number 69 are the numbers 1, 3, 23 and 69, then 69 can be obtained in two ways: 69=1·69 and 69=3·23. Considering that , we get two systems of equations, by solving which we can find the required numbers:
The first system has a solution, and the second system has a solution.
Answer: .
Task 5.
Solution. Let's write the equation in the form
.
Let's factorize the left side of the equation. We get
.
The product of two integers can equal 1 only in two cases: if they are both equal to 1 or -1. We get two systems:
The first system has a solution x=2, y=2, and the second system has a solution x=0, y=0.
Answer: .
Task 6. Solve the equation in whole numbers
.
Solution. Let us write this equation in the form
Let's factorize the left side of the equation using the grouping method, we get
.
The product of two integers can equal 7 in following cases:
7=1· 7=7·1=-1·(-7)=-7·(-1). Thus, we get four systems:
Or, or, or.
The solution to the first system is a pair of numbers x = - 5, y = - 6. Solving the second system, we get x = 13, y = 6. For the third system, the solution is the numbers x = 5, y = 6. The fourth system has a solution x = - 13, y = - 6.
Task 7. Prove that the equation ( x - y) 3 + (y - z) 3 + (z - x) 3 = 30 not
has solutions in integers.
Solution. 1) Let's factorize the left side of the equation and divide both sides of the equation by 3, resulting in the following equation:
(x - y)(y - z)(z - x) = 10…………………………(2)
2) The divisors of 10 are the numbers ±1, ±2, ±5, ±10. Note also that the sum of the factors on the left side of equation (2) is equal to 0. It is easy to check that the sum of any three numbers from the set of divisors of the number 10, giving the product 10, will not equal 0. Consequently, the original equation has no solutions in integers.
Task 8. Solve the equation: x 2 - y 2 = 3 in whole numbers.
Solution:
1. apply the abbreviated multiplication formula x 2 - y 2 = (x-y)(x+y)=3
2. find the divisors of the number 3 = -1;-3;1;3
3. This equation is equivalent to a set of 4 systems:
X-y=1 2x=4 x=2, y=1
X-y=3 x=2, y=-1
X-y=-3 x=-2, y=1
X-y=-1 x=-2, y=-1
Answer: (2;1), (2;-1), (-2;1), (-2,-1)
2.4 Residual method.
Problem 9 .Solve the equation: x 2 + xy = 10
Solution:
1. Express the variable y through x: y= 10's 2
Y = - X
2. Fraction will be integer if x Є ±1;±2; ±5;±10
3. Find 8 values u.
If x=-1, then y=-9 x=-5, then y=3
X=1, then y=9 x=5, then y=-3
X=-2, then y=-3 x=-10, then y=9
X=2, then y=3 x=10, then y=-9
Problem 10. Solve the equation in whole numbers:
2x 2 -2xy +9x+y=2
Solution:
Let us express from the equation the unknown that is included in it only to the first degree - to in this case y:
2x 2 +9x-2=2xy-y
Y = 
Let's select the whole part of a fraction using the rule of dividing a polynomial by a polynomial by an “angle”. We get:
Therefore, the difference 2x-1 can only take the values -3,-1,1,3.
It remains to go through these four cases.
Answer : (1;9), (2;8), (0;2), (-1;3)
2. Exam level tasks
Having considered several ways to solve first-degree equations with two variables in integers, we noticed that the method of factorization and the method of remainders are most often used.
The equations that are given in Unified State Exam options-2011, are mainly solved by the residual method.
1. Solve the equation in natural numbers: , where m>n
Solution:
Let's express the variable n via variable T
(y+10) 2< 6 -2 ≤ у+10 ≤ 2 -12 ≤ у ≤ -8
(y+6) 2< 5 -2 ≤ у+6 ≤ 2 -8 ≤ у ≤ -4 у=-8
Answer: (12; -8)
Conclusion.
Solution various types equations is one of the content lines of the school mathematics course, but methods for solving equations with several unknowns are practically not considered. At the same time, solving equations of several unknowns in integers is one of the oldest mathematical problems. Most methods for solving such equations are based on the theory of divisibility of integers, interest in which is currently determined by the rapid development of information technology. In this regard, it will be interesting for high school students to get acquainted with methods for solving some equations in integers, especially since the Olympiads different levels very often tasks are offered that involve solving some equation in integers, and this year such equations are also included in the Unified State Exam materials.
In our work, we considered only indeterminate equations of the first and second degrees. Equations of the first degree, as we have seen, are solved quite simply. We have identified the types of such equations and algorithms for solving them. It was also found general solution such equations.
With equations of the second degree it is more difficult, so we considered only special cases: the Pythagorean theorem and cases when one part of the equation has the form of a product, and the second is factorized.
Great mathematicians study equations of third and higher degrees because their solutions are too complex and cumbersome
In the future, we plan to deepen our research in the study of equations with several variables that are used in solving problems
Literature.
1. Berezin V.N. Collection of problems for elective and extracurricular activities in mathematics. Moscow "Enlightenment" 1985
2. Galkin E.G. Non-standard problems in mathematics. Chelyabinsk “Vzglyad” 2004
3. Galkin E.G. Problems with integers. Chelyabinsk “Vzglyad” 2004
4. Glazer E.I. History of mathematics at school. Moscow “Enlightenment” 1983
5. Mordkovich A.G. Algebra and beginning of analysis grades 10-11. Moscow 2003
6. Mathematics. Unified State Exam 2010. Federal Institute
pedagogical measurements.
7. Sharygin I.F. Optional course in mathematics. Solution
tasks. Moscow 1986
Problems with integer unknowns
Pavlovskaya Nina Mikhailovna,
mathematics teacher MBOU "Secondary School No. 92
Kemerovo
Algebraic equations or systems algebraic equations with integer coefficients having a number of unknowns exceeding the number of equations, and for which integer or rational decisions got the name diophantine equations .
The problem of solving equations in integers has been completely solved only for equations with one unknown, for equations of the first degree and for equations of the second degree with two unknowns. For equations above the second degree with two or more unknowns, even the task of proving the existence of integer solutions is difficult. Moreover, it has been proven that there is no single algorithm that allows solving arbitrary Diophantine equations in integers in a finite number of steps.
- The simplest Diophantine equations are equations of the form
ax + by = c , a ≠ 0; b ≠ 0
If c = 0 , then the solution is obvious: x = 0, y = 0.
If s ≠ 0 , and the solution (X 0 ; at 0 ) , then an integer
ax 0 + by 0 divided by d = (a ; b) , That's why With must also be divisible by a common divisor a and b .
For example: 3x + 6y = 5 has no integer solutions, since (3; 6) = 3, and c = 5 is not divisible by 3 without a remainder.
- If the equation ax + by = c has a solution (X 0 ; at 0 ) , And (a ; b) = 1 , then all solutions of the equation are given by the formulas x = x 0 +bn; y = y 0 –an, where n is any integer solution.
For example: 3x + 5y = 13, (3; 5) = 1, which means the equation has infinitely many solutions, X 0 =1; at 0 =2
Fermat's great theorem states: an equation of the form has no solutions in natural numbers.
This theorem was formulated by the Italian mathematician Pierre Fermat more than 300 years ago, and was proven only in 1993.
Factorization method .
1) Solve the equation in integers
x + y = xy.
Solution. Let's write the equation in the form
(x - 1)(y - 1) = 1.
The product of two integers can equal 1 only if both of them are equal to 1. That is, the original equation is equivalent to the aggregate
with solutions (0,0) and (2,2).
2. Solve the equation in whole numbers:
3x² + 4xy – 7y²= 13.
Solution: 3x² - 3xy + 7xy – 7y²= 13,
3x(x – y) + 7y(x – y) = 13,
(x – y)(3x + 7y) = 13.
Since 13 has integer divisors ±1 and ±13,
1. x – y = 1, 7x – 7y = 7, x = 2,
3x + 7y = 13; 3x + 7y = 13; whence y = 1
2. x – y = 13, 7x – 7y = 91, x = 9.2,
3x + 7y= 1; 3x + 7y = 1; where y = - 3.8.
3 . x – y = -1, 7x – 7y = -7, x = -2,
3x + 7y= -13; 3x + 7y = -13; whence y = -1.
4. x – y = -13, 7x – 7y = -91, x = -9.2,
3x + 7y= -1; 3x +7y= -1; whence y = 3.8.
Therefore, the equation has two solutions in integers: (2;1) and (-2;-1)
3 . Solve the equation in whole numbers:
9x² + 4x – xy + 3y = 88.
Solution: 9x² + 4x – 88 = xy – 3y,
9x² + 4x – 88 = y(x – 3)
since 5 has integer divisors ± 1 and ± 5, then
The text of the work is posted without images and formulas.
Full version work is available in the "Work Files" tab in PDF format
Object of study.
The research concerns one of the most interesting sections of number theory - solving equations in integers.
Subject of research.
Solving in integer algebraic equations with integer coefficients and more than one unknown is one of the most difficult and ancient mathematical problems and is not sufficiently presented in the school mathematics course. In my work I will present enough full analysis equations in integers, classification of these equations according to methods for solving them, description of algorithms for solving them, as well as practical examples applying each method to solve equations in integers.
Target.
Learn how to solve equations in integers.
Tasks:
Study educational and reference literature;
Collect theoretical material on how to solve equations;
Analyze algorithms for solving equations of this type;
Describe solutions;
Consider examples of solving equations using these methods.
Hypothesis:
Having encountered equations in integers in Olympiad tasks, I assumed that the difficulties in solving them were due to the fact that not all methods for solving them were known to me.
Relevance:
While solving sample versions of Unified State Exam tasks, I noticed that there are often tasks for solving equations of the first and second degrees in integers. In addition, Olympiad tasks different levels also contain integer equations or problems that are solved using integer equation solving skills. The importance of knowing how to solve equations in integers determines the relevance of my research.
Research methods
Theoretical analysis and generalization of scientific literature information about equations in integers.
Classification of equations in integers according to methods for solving them.
Analysis and generalization of methods for solving equations in integers.
Research results
The work describes methods for solving equations, considers the theoretical material of Fermat's theorem, Pythagorean theorem, and Euclid's algorithm, and presents examples of solutions to problems and equations of various levels of complexity.
2. History of equations in integers
Diophantus was a scientist and algebraist of Ancient Greece; according to some sources, he lived until 364 AD. e. He specialized in solving problems in integers. This is where the name Diophantine equations comes from. The most famous, solved by Diophantus, is the problem of “decomposition into two squares.” Its equivalent is the well-known Pythagorean theorem. The life and work of Diophantus took place in Alexandria, he collected and solved known problems and came up with new ones. He later combined them in a great work called Arithmetic. Of the thirteen books that made up the Arithmetic, only six survived into the Middle Ages and became a source of inspiration for Renaissance mathematicians. Diophantus' Arithmetic is a collection of problems, each including a solution and the necessary explanation. The meeting includes a variety of tasks, and their solution is often highest degree witty. Diophantus is only interested in positive integers and rational solutions. Irrational decisions he calls them “impossible” and carefully selects the coefficients so that the desired positive, rational solutions are obtained.
Fermat's theorem is used to solve equations in integers. The history of the proof of which is quite interesting. Over full proof Many outstanding mathematicians worked on the Great Theorem, and these efforts led to many results modern theory numbers. It is believed that the theorem ranks first in terms of the number of incorrect proofs.
The remarkable French mathematician Pierre Fermat stated that the equation for integer n ≥ 3 has no solutions in positive integers x, y, z (xyz = 0 is excluded by the positivity of x, y, z. For the case n = 3, this theorem was tried in the 10th century proved by the Central Asian mathematician al-Khojandi, but his proof was not preserved. Somewhat later, Fermat himself published a proof of a special case for n = 4.
Euler in 1770 proved the theorem for the case n = 3, Dirichlet and Legendre in 1825 - for n = 5, Lame - for n = 7. Kummer showed that the theorem is true for all primes n less than 100, with the possible exception of 37, 59, 67.
In the 1980s there appeared new approach to solving the problem. From Mordell's conjecture, proven by Faltings in 1983, it follows that the equation
for n > 3 can have only a finite number of relatively simple solutions.
The last, but most important, step in proving the theorem was taken in September 1994 by Wiles. His 130-page proof was published in the journal Annals of Mathematics. The proof is based on the assumption of the German mathematician Gerhard Frey that Fermat's Last Theorem is a consequence of the Taniyama-Shimura conjecture (this assumption was proved by Ken Ribet with the participation of J.-P. Serres). Wiles published the first version of his proof in 1993 (after 7 years of hard work), but a serious gap soon emerged; With the help of Richard Lawrence Taylor, the gap was quickly closed. The final version was published in 1995. 15 March 2016 Andrew Wiles receives the Abel Prize. Currently, the premium is 6 million Norwegian kroner, that is, approximately 50 million rubles. According to Wiles, the award came as a “complete surprise” to him.
3. Linear equations in integers
Linear equations are the simplest of all Diophantine equations.
An equation of the form ax=b, where a and b are some numbers and x is an unknown variable, is called a linear equation with one unknown. Here you need to find only integer solutions to the equation. It can be noted that if a ≠ 0, then the equation will have an integer solution only if b is completely divisible by a and this solution is x = b/ph. If a=0, then the equation will have an integer solution when b=0 and in this case x is any number.
because 12 is divisible by 4, then
Because a=o and b=0, then x is any number
Because 7 is not completely divisible by 10, then there are no solutions.
4. Method of enumerating options.
In the method of enumerating options, it is necessary to take into account the signs of divisibility of numbers, consider all possible options finite search equality. This method can be used to solve these problems:
№1 Find the set of all pairs of natural numbers that are a solution to the equation 49x+69y=602
We express from the equation x =,
Because x and y are natural numbers, then x = ≥ 1, multiply the entire equation by 49 to get rid of the denominator:
Move 602 to the left:
51y ≤ 553, express y, y= 10
A complete search of options shows that the natural solutions to the equation are x=5, y=7.
Answer:(5.7).-
№2 Solve the problem
From the numbers 2, 4, 7, you should create a three-digit number in which not a single digit can be repeated more than twice.
Let's find the number of all three-digit numbers that begin with the number 2: (224, 242, 227, 272, 247, 274, 244, 277) - there are 8 of them.
Similarly, we find all three-digit numbers starting with numbers 4 and 7: (442, 424, 422, 447, 474, 427, 472, 477).
(772, 774, 727, 747, 722, 744, 724, 742) - there are also 8 numbers each. It's only the 24th.
Answer: 24th.
5. Continued fraction and Euclidean algorithm
A continued fraction is the expression common fraction in the form
where q 1 is an integer, and q 2, ..., qn are natural numbers. This expression is called a continued (finite continued) fraction. There are finite and infinite continued fractions.
For rational numbers the continued fraction has a finite form. In addition, the sequence a i is exactly the sequence of quotients that is obtained by applying the Euclidean algorithm to the numerator and denominator of a fraction.
Solving equations with continued fractions, I compiled a general algorithm for this method solving equations in integers.
Algorithm
1) Compose the ratio of coefficients for unknowns in the form of a fraction
2) Convert the expression to an improper fraction
3) Select the whole part improper fraction
4) Replace a proper fraction with an equal fraction
5) Do 3.4 with the resulting improper fraction in the denominator
6) Repeat 5 until the final result
7) In the resulting expression, discard the last link of the continued fraction, turn the resulting new continued fraction into a simple one and subtract it from the original fraction.
Example№1 Solve the equation 127x- 52y+ 1 = 0 in integers
Let us transform the ratio of the coefficients for the unknowns.
First of all, let's select the whole part of the improper fraction; = 2 +
We replace the proper fraction with an equal fraction.
From = 2+
Let's do the same transformations with the improper fraction obtained in the denominator.
Now the original fraction will take the form: . Repeating the same reasoning for the fraction we get. Isolating the whole part of the improper fraction, we come to the final result:
We have obtained an expression called a finite continued fraction. Having discarded the last link of this continued fraction - one fifth, we turn the resulting new continued fraction into a simple one and subtract it from the original fraction:
Let us reduce the resulting expression to a common denominator and discard it.
Where does 127∙9-52∙22+1=0 come from. From a comparison of the resulting equality with the equation 127x- 52y+1 = 0 it follows that then x= 9, y= 22 is the solution to the original equation, and according to the theorem, all its solutions will be contained in the progressions x= 9+ 52t, y= 22+ 127t , where t=(0; ±1; ±2…..). The result obtained suggests that in the general case, to find a solution to the equation ax+by+c=0, it is necessary to expand the ratio of the coefficients of the unknowns into a continued fraction , discard its last link and do the calculations, similar topics which were given above.
To prove this assumption, we will need some properties of continued fractions.
Let's consider an irreducible fraction. Let us denote by q 1 the quotient and by r 2 the remainder of division of a by b. Then we get:
Then b=q 2 r 2 +r 3 ,
Exactly the same
r 2 =q 3 r 3 +r 4 , ;
r 3 =q 4 r 4 +r 5 ,;
………………………………..
The quantities q 1, q 2,... are called incomplete quotients. The above process of formation of incomplete quotients is called Euclidean algorithm. The remainders from division r 2 , r 3 ,…satisfy the inequalities
those. form a series of decreasing non-negative numbers.
Example No. 2 Solve the equation 170x+190y=3000 in integers
After reducing by 10 the equation looks like this:
To find a particular solution, we use the decomposition of a fraction into a continued fraction
By collapsing the penultimate fraction that matches it into an ordinary fraction
A particular solution to this equation has the form
X 0 = (-1)4300∙9=2700, y 0 =(-1)5300∙8=-2400,
and the general one is given by the formula
x=2700-19k, y= -2400+17k.
from which we obtain the condition for the parameter k
Those. k=142, x=2, y=14. .
6. Factorization method
The method of enumerating options is an inconvenient method, since there are cases when it is impossible to find complete solutions by enumeration, since there are an infinite number of such solutions. The factorization method is a very interesting technique and it is found both in elementary and higher mathematics.
The essence is the identity transformation. The meaning of any identity transformation is a recording of an expression in a different form while preserving its essence. Let's look at examples of using this method.
№1 Solve the equation in integers y 3 -x 3 = 91.
Using abbreviated multiplication formulas, we factorize the right side of the equation:
(y - x)(y 2 + xy + x 2) = 91
We write down all the divisors of the number 91: ± 1; ± 7; ± 13; ± 91
We note that for any integer x and y the number
y 2 + yx + x 2 ≥ y 2 - 2|y||x| + x 2 = (|y| - |x|) 2 ≥ 0,
therefore, both factors on the left side of the equation must be positive. Then the original equation is equivalent to a set of systems of equations:
Having solved the systems, we select those roots that are integers.
We obtain solutions to the original equation: (5; 6), (-6; -5); (-3; 4),(-4; 3).
Answer: (5; 6); (-6; -5); (-3; 4); (-4;3).
№2 Find all pairs of natural numbers satisfying the equation x 2 -y 2 = 69
Let's factorize the left side of the equation and write the equation in the form
Because The divisors of the number 69 are the numbers 1, 3, 23 and 69, then 69 can be obtained in two ways: 69=1·69 and 69=3·23. Considering that x-y > 0, we get two systems of equations, solving which we can find the required numbers:
By expressing one variable and substituting it into the second equation, we find the roots of the equations. The first system has a solution x=35;y=34, and the second system has a solution x=13, y=10.
Answer: (35; 34), (13; 10).
№3 Solve the equation x + y = xy in integers:
Let's write the equation in the form
Let's factorize the left side of the equation. We get
The product of two integers can equal 1 only in two cases: if they are both equal to 1 or -1. We get two systems:
The first system has a solution x=2, y=2, and the second system has a solution x=0, y=0. Answer: (2; 2), (0; 0).
№4 Prove that the equation (x - y) 3 + (y - z) 3 + (z - x) 3 = 30 has no solutions in integers.
Let's factorize the left side of the equation and divide both sides of the equation by 3, resulting in the following equation:
(x - y)(y - z)(z - x) = 10
The divisors of 10 are the numbers ±1, ±2, ±5, ±10. Note also that the sum of the factors on the left side of the equation is equal to 0. It is easy to check that the sum of any three numbers from the set of divisors of the number 10, giving the product 10, will not equal 0. Consequently, the original equation has no solutions in integers.
7. Residual method
The main task of the method is to find the remainder when dividing both sides of the equation by an integer, based on the results obtained. Often the information obtained reduces the possibilities of solution sets for the equation. Let's look at examples:
№1 Prove that the equation x 2 = 3y + 2 has no integer solutions.
Proof.
Consider the case when x, y ∈ N. Consider the remainder when both sides are divided by 3. The right side of the equation gives a remainder of 2 when divided by 3 for any value of y. The left side, which is the square of a natural number, when divided by 3 always gives a remainder of 0 or 1. Based on this, we find that there is no solution to this equation in natural numbers.
Let's consider the case when one of the numbers is 0. Then, obviously, there are no solutions in integers.
The case when y is a negative integer has no solutions, because right side will be negative, and the left one will be positive.
The case when x is a negative integer also has no solutions, because falls under one of the previously considered cases due to the fact that (-x) 2 = (x) 2.
It turns out that the indicated equation has no solutions in integers, which is what needed to be proven.
№2 Solve in whole numbers 3 X = 1 + y 2 .
It is not difficult to notice that (0; 0) is the solution to this equation. It remains to prove that the equation has no other integer roots.
Let's consider the cases:
1) If x∈N, y∈N, then 3 is divisible by three without a remainder, and 1 + y 2 when divided by 3 gives
the remainder is either 1 or 2. Therefore, equality for natural
values x, y is impossible.
2) If x is a negative integer, y∈Z, then 0< 3 х < 1, а 1 + y 2 ≥ 0 и
equality is also impossible. Therefore, (0; 0) is the only
Answer: (0; 0).
№3 Solve equation 2x 2 -2xy+9x+y=2 in integers:
Let us express from the equation the unknown that is included in it only to the first degree, that is, the variable y:
2x 2 +9x-2=2xy-y, where from
Let us select the whole part of a fraction using the rule of dividing a polynomial by a polynomial by an “angle”. We get:
Obviously, the difference 2x-1 can only take the values -3, -1, 1 and 3.
It remains to go through these four cases, as a result of which we obtain the solutions: (1;9), (2;8), (0;2), (-1;3)
Answer: (1;9), (2;8), (0;2), (-1;3)
8. An example of solving equations with two variables in integers as square with respect to one of the variables
№1 Solve the equation 5x in whole numbers 2 +5у 2 + 8xy+2y-2x +2=0
This equation can be solved by factorization, but this method, when applied to this equation, is quite labor-intensive. Let's consider a more rational way.
Let's write the equation in quadratic form with respect to the variable x:
5x 2 +(8y-2)x+5y 2 +2y+2=0
We find its roots.
This equation has a solution if and only if the discriminant
of this equation is equal to zero, i.e. - 9(y+1) 2 =0, hence y= - 1.
If y= -1, then x= 1.
Answer: (1; - 1).
9.An example of solving problems using equations in integers.
№ 1. Solve the equation in natural numbers : where n>m
Let's express the variable n through the variable m:
Let's find the divisors of the number 625: this is 1; 5; 25; 125; 625
1) if m-25 =1, then m=26, n=25+625=650
2) m-25 =5, then m=30, n=150
3) m-25 =25, then m=50, n=50
4) m-25 =125, then m=150, n=30
5) m-25 =625, then m=650, n=26
Answer: m=150, n=30
№ 2. Solve the equation in natural numbers: mn +25 = 4m
Solution: mn +25 = 4m
1) express the variable 4m in terms of n:
2) find the natural divisors of the number 25: this is 1; 5; 25
if 4-n =1, then n=3, m=25
4-n=5, then n=-1, m=5; 4-n =25, then n=-21, m=1 (extraneous roots)
Answer: (25;3)
In addition to tasks to solve an equation in integers, there are tasks to prove the fact that the equation does not have integer roots.
When solving such problems, it is necessary to remember the following properties of divisibility:
1) If n Z; n is divisible by 2, then n = 2k, k ∈ Z.
2) If n ∈ Z; n is not divisible by 2, then n = 2k+1, k ∈ Z.
3) If n ∈ Z; n is divisible by 3, then n = 3k, k ∈ Z.
4) If n ∈ Z; n is not divisible by 3, then n = 3k±1, k ∈ Z.
5) If n ∈ Z; n is not divisible by 4, then n = 4k+1; n = 4k+2; n = 4k+3. k ∈ Z.
6) If n ∈ Z; n(n+1) is divisible by 2, then n (n+1)(n+2) is divisible by 2;3;6.
7)n; n+1 are relatively prime.
№3 Prove that the equation x 2 - 3y = 17 has no integer solutions.
Proof:
Let x; y - solutions of the equation
x 2 = 3(y+6)-1 Because. y ∈ Z then y+6 ∈ Z, which means 3(y+6) is divisible by 3, therefore, 3(y+6)-1 is not divisible by 3, therefore, x 2 is not divisible by 3, therefore, x is not divisible by 3, which means x = 3k±1, k ∈ Z.
Let's substitute this into the original equation.
We got a contradiction. This means that the equation has no entire solutions, which is what needed to be proven.
10.Pica formula
The Pieck formula was discovered by the Austrian mathematician Georg Pieck in 1899. The formula is related to equations in integers in that only integer nodes are taken from polygons, just like integers in the equations.
Using this formula, you can find the area of a figure constructed on a sheet of paper in a cage (triangle, square, trapezoid, rectangle, polygon).
In this formula we will find whole points inside the polygon and on its border.
In the problems that will be on the Unified State Exam there is a whole group of tasks in which a polygon is given, constructed on a sheet of paper in a square, and the question is about finding the area. The cell scale is one square centimeter.
Example No. 1
M - number of nodes on the border of the triangle (on the sides and vertices)
N is the number of nodes inside the triangle.
*By “nodes” we mean the intersection of lines. Let's find the area of the triangle:
Let's mark the nodes:
M = 15 (indicated in red)
N=34 (in blue)
Example No. 2
Let's find the area of the polygon: Mark the nodes:
M = 14 (indicated in red)
N=43 (in blue)
12.Descent method
One of the methods for solving equations in integers - the descent method - is based on Fermat's theorem.
The descent method is a method that consists in constructing one solution to an infinite sequence of solutions with infinitely decreasing positive z.
Let's consider the algorithm of this method using the example of solving a specific equation.
Example 1. Solve the equation in integers 5x + 8y = 39.
1) Let’s choose the unknown that has the smallest coefficient (in our case it’s x), and express it through another unknown:
2) Select the integer part: Obviously, x will be an integer if the expression turns out to be an integer, which, in turn, will occur when the number 4 - 3y is divisible by 5 without a remainder.
3) Let's introduce an additional integer variable z as follows: 4 -3y = 5z. As a result, we obtain an equation of the same type as the original one, but with smaller coefficients.
4) We solve it with respect to the variable y, reasoning exactly as in points 1, 2: Selecting the whole part, we get:
5) Reasoning similarly to the previous one, we introduce a new variable u: 3u = 1 - 2z.
6) Express the unknown with the smallest coefficient, in this case the variable z: . Requiring that it be an integer, we get: 1 - u = 2v, whence u = 1 - 2v. There are no more fractions, the descent is completed (we continue the process until there are no fractions left in the expression for the next variable).
7) Now you need to “go up”. Let us express through the variable v first z, then y and then x:
8) The formulas x = 3+8v and y = 3 - 5v, where v is an arbitrary integer, represent the general solution to the original equation in integers.
Thus, the descent method involves first sequentially expressing one variable in terms of another until there are no fractions left in the representation of the variable, and then sequentially “ascending” along the chain of equalities to obtain a general solution to the equation.
12.Conclusion
As a result of the study, the hypothesis was confirmed that the difficulties in solving equations in integers are due to the fact that not all methods for solving them were known to me. During my research, I was able to find and describe little-known methods solutions of equations in integers, illustrate them with examples. The results of my research can be useful to all students interested in mathematics.
13.Bibliography
Book resources:
1. N. Ya. Vilenkin et al., Algebra and mathematical analysis / 10th grade, 11th grade // M., “Enlightenment”, 1998;
2. A.F. Ivanov et al., Mathematics. Educational and training materials for preparing for the exam // Voronezh, GOUVPO VSTU, 2007
3. A. O. Gelfond, Mathematics, number theory // Solving equations in integers // LIBROKOM Book House
Internet resources:
4. Demo options control measuring materials of a single state exam in mathematics http://fipi.ru/
5. Examples of solutions to equations in integers http://reshuege.ru
6. Examples of solutions to equations in integers http://mat-ege.ru
7. History of Diophantine equations http://www.goldenmuseum.com/1612Hilbert_rus.html
8. History of Diophantus http://nenuda.ru/%D1%83%D1%80%D0%B0%D0%B2%D0%BD%D0%B5%D0%BD%D0%B8%D1%8F-% D1%81-%D0%B4%D0%B2%D1%83%D0%BC%D1%8F-%D0%BD%D0%B5%D0%B8%D0%B7%D0%B2%D0%B5% D1%81%D1%82%D0%BD%D1%8B%D0%BC%D0%B8-%D0%B2-%D1%86%D0%B5%D0%BB%D1%8B%D1%85- %D1%87%D0%B8%D1%81%D0%BB%D0%B0%D1%85.htm
9. History of Diophantine equations http://dok.opredelim.com/docs/index-1732.html
10. History of Diophantus http://www.studfiles.ru/preview/4518769/
There are many paths leading from the edge of the forest into the thicket. They are tortuous, they converge, diverge again and intersect with each other again. While walking, you can only notice the abundance of these paths, walk along some of them and trace their direction into the depths of the forest. To seriously study the forest, you need to follow the paths until they are visible at all among the dry pine needles and bushes.
Therefore, I wanted to write a project that can be considered as a description of one of the possible walks along the edge of modern mathematics.
The world around us, needs national economy, and often everyday chores present a person with more and more new tasks, the solution to which is not always obvious. Sometimes a particular question has many possible answers, which causes difficulties in solving the tasks. How to choose the right and optimal option?
The solution of indeterminate equations is directly related to this issue. Such equations, containing two or more variables, for which it is necessary to find all integer or natural solutions, have been considered since ancient times. For example, the Greek mathematician Pythagoras (IV century BC). the Alexandrian mathematician Diophantus (II-III century AD) and the best mathematicians of a closer era to us - P. Fermat (XVII century), L. Euler (XVIII century), J. L. Lagrange (XVIII century) and others.
Participating in the Russian correspondence competition > Obninsk, International competition- game > and Ural Olympiad Federal District I often encounter such problems. This is due to the fact that their solution is creative. The problems that arise when solving equations in integers are caused both by complexity and by the fact that little time is devoted to them in school.
Diophantus presents one of the most difficult mysteries in the history of science. We do not know the time when he lived, nor his predecessors who would have worked in the same field. His works are like a sparkling fire in the midst of impenetrable darkness.
The period of time when Diophantus could have lived is half a millennium! The lower bound is determined without difficulty: in his book on polygonal numbers, Diophantus repeatedly mentions the mathematician Hypsicles of Alexandria, who lived in the middle of the 2nd century. BC e.
On the other hand, in the comments of Theon of Alexandria to the famous astronomer Ptolemy there is an excerpt from the work of Diophantus. Theon lived in the middle of the 4th century. n. e. This determines the upper bound of this interval. So, 500 years!
The French historian of science Paul Tannry, editor of the most complete text of Diophantus, tried to narrow this gap. In the Escurial library he found excerpts from a letter from Michael Psellus, a Byzantine scholar of the 11th century. , where it is said that the most learned Anatoly, after collecting the most essential parts of this science, we are talking about the introduction of degrees of the unknown and about their (designation), dedicated them to his friend Diophantus. Anatoly of Alexandria actually composed >, excerpts of which are cited in the extant works of Iamblichus and Eusenius. But Anatoly lived in Alexandria in the middle of the 111th century BC. e and even more precisely - until 270, when he became the bishop of Laodacia. This means that his friendship with Diophantus, whom everyone calls Alexandria, must have taken place before this. So, if the famous Alexandrian mathematician and Anatoly's friend named Diophantus are one person, then the time of Diophantus's life is the middle of the 111th century AD.
But the place of residence of Diophantus is well known - Alexandria, the center of scientific thought and the Hellenistic world.
One of the epigrams of the Palatine Anthology has survived to this day:
The ashes of Diophantus rest in the tomb: marvel at it - and the stone
The age of the deceased will speak through his wise art.
By the will of the gods, he lived a sixth of his life as a child.
And I met half past five with fluff on my cheeks.
It was only the seventh day when he became engaged to his girlfriend.
After spending five years with her, the sage waited for his son.
His father's beloved son lived only half his life.
He was taken from his father by his early grave.
Twice for two years the parent mourned severe grief.
Here I saw the limit of my sad life.
Using modern methods solutions to equations can be calculated how many years Diophantus lived.
Let Diophantus live x years. Let's create and solve the equation:
Let's multiply the equation by 84 to get rid of fractions:
Thus, Diophantus lived 84 years.
The most mysterious is the work of Diophantus. Six of the thirteen books that were combined into > have reached us; the style and content of these books differ sharply from the classical ancient works on number theory and algebra, examples of which we know from > Euclid, his >, lemmas from the works of Archimedes and Apollonius. > was undoubtedly the result of numerous studies that remained completely unknown.
We can only guess about its roots, and marvel at the richness and beauty of its methods and results.
> Diophanta is a collection of problems (189 in total), each of which has a solution. The problems in it are carefully selected and serve to illustrate very specific, strictly thought out methods. As was customary in ancient times, the methods are not formulated in general view, but are repeated to solve similar problems.
A unique biography of Diophantus is reliably known, which, according to legend, was carved on his tombstone and represented a puzzle task:
This puzzle serves as an example of the problems that Diophantus solved. He specialized in solving problems in integers. Such problems are currently known as Diophantine problems.
The study of Diophantine equations is usually associated with great difficulties.
In 1900, at the World Congress of Mathematicians in Paris, one of the world's greatest mathematicians, David Hilbert, identified 23 problems from various areas mathematics. One of these problems was the problem of solving Diophantine equations. The problem was the following: is it possible to solve an equation with an arbitrary number of unknowns and integer coefficients in a certain way - using an algorithm. The task is as follows: for a given equation, you need to find all the integer or natural values of the variables included in the equation, at which it turns into a true equality. Diophantus came up with many different solutions for such equations. Due to the infinite variety of Diophantine equations, there is no general algorithm for solving them, and for almost each equation one has to invent an individual technique.
A Diophantine equation of the 1st degree or a linear Diophantine equation with two unknowns is an equation of the form: ax+by=c, where a,b,c are integers, GCD(a,b)=1.
I will give the formulations of theorems on the basis of which an algorithm for solving indeterminate first-degree equations of two variables in integers can be compiled.
Theorem 1. If in an equation, then the equation has at least one solution.
Proof:
We can assume that a >0. Having solved the equation for x, we get: x = c-vua. I will prove that if in this formula instead of y we substitute all natural numbers less than a and 0, i.e. the numbers 0;1;2;3;. ;a-1, and each time you perform division, then all a remainders will be different. Indeed, instead of y, I will substitute the numbers m1 and m2, smaller than a. As a result, I will get two fractions: c-bm1a and c-bm2a. Having performed the division and denoted the incomplete quotients by q1 and q2, and the remainders by r1 and r2, I will find с-вm1а=q1+ r1а, с-вm2а= q2+ r2а.
I will assume that the remainders r1 and r2 are equal. Then, subtracting the second from the first equality, I get: c-bm1a- c-bm2a = q1-q2, or b(m1 - m2)a = q1-q2.
Since q1-q2 is an integer, then left side must be intact. Therefore, bm1 - m2 must be divisible by a, i.e., the difference of two natural numbers, each of which is less than a, must be divisible by a, which is impossible. This means that the remainders r1 and r2 are equal. That is, all residues are different.
That. I received a of various balances less than a. But the distinct a of natural numbers not exceeding a are the numbers 0;1;2;3;. ;a-1. Consequently, among the remainders there will certainly be one and only one equal to zero. The value of y, the substitution of which into the expression (c-vu)a gives a remainder of 0, and turns x=(c-vu)a into an integer. Q.E.D.
Theorem 2. If in the equation, and c is not divisible by, then the equation has no integer solutions.
Proof:
Let d=GCD(a;b), so that a=md, b=nd, where m and n are integers. Then the equation will take the form: mdх+ ndу=с, or d(mх+ nу)=с.
Assuming that there are integers x and y that satisfy the equation, I find that the coefficient c is divisible by d. The resulting contradiction proves the theorem.
Theorem 3. If in the equation, and, then it is equivalent to the equation in which.
Theorem 4. If in an equation, then all integer solutions to this equation are contained in the formulas:
where x0, y0 is an integer solution to the equation, is any integer.
The formulated theorems make it possible to construct the following algorithm for solving an equation of the form in integers.
1. Find the greatest common divisor of the numbers a and b; if c is not divisible by, then the equation has no integer solutions; if and then
2. Divide the equation term by term, obtaining an equation in which.
3. Find an integer solution (x0, y0) of the equation by representing 1 as a linear combination of numbers and;
4. Create a general formula for integer solutions to this equation, where x0, y0 is an integer solution to the equation, and is any integer.
2. 1 DESCENT METHOD
Many > are based on methods for solving uncertain equations. For example, a trick involving guessing the date of birth.
Invite your friend to guess his birthday by the sum of numbers equal to the product of his date of birth by 12 and the number of the month of birth by 31.
In order to guess the birthday of your friend you need to solve the equation: 12x + 31y = A.
Let you be given the number 380, i.e. we have the equation 12x + 31y = 380. In order to find x and y, you can reason like this: the number 12x + 24y is divisible by 12, therefore, according to the properties of divisibility (Theorem 4.4), the number 7y and 380 must have the same remainder when divided by 12. The number 380 when divided by 12 gives a remainder of 8, therefore 7y when divided by 12 must also leave a remainder of 8, and since y is the number of the month, then 1
The equation we solved is a 1st degree Diophantine equation with two unknowns. To solve such equations, the so-called descent method can be used. I will consider the algorithm of this method using the specific equation 5x + 8y = 39.
1. I will choose the unknown that has the smallest coefficient (in our case it is x), and express it through another unknown:. I'll highlight the whole part: Obviously, x will be an integer if the expression turns out to be an integer, which, in turn, will be the case when the number 4 - 3y is divisible by 5 without a remainder.
2. I will introduce an additional integer variable z as follows: 4 - 3y = 5z. As a result, I will get an equation of the same type as the original one, but with smaller coefficients. I will solve it with respect to the variable y:. Selecting the whole part, I get:
Reasoning similarly to the previous one, I introduce a new variable u: 3u = 1 - 2z.
3. I will express the unknown with the smallest coefficient, in this case the variable z: =. Requiring that it be an integer, I get: 1 - u = 2v, whence u = 1 - 2v. There are no more fractions, the descent is complete.
4. Now you need >. I will express through the variable v first z, then y and then x: z = = = 3v - 1; = 3 - 5v.
5. The formulas x = 3+8v and y = 3 - 5v, where v is an arbitrary integer, represent the general solution to the original equation in integers.
Comment. Thus, the descent method involves first sequentially expressing one variable in terms of another until there are no fractions left in the representation of the variable, and then sequentially along a chain of equalities to obtain a general solution to the equation.
2. 2 SURVEY METHOD
Rabbits and pheasants sit in a cage; they have 18 legs in total. Find out how many of both are in the cell?
Let me create an equation with two unknowns, in which x is the number of rabbits, and y is the number of pheasants:
4x + 2y = 18, or 2x + y = 9.
Answer. 1) 1 rabbit and 7 pheasants; 2) 2 rabbits and 5 pheasants; 3) 3 rabbits and 3 pheasants; 4) 4 rabbits and 1 pheasant.
1. PRACTICAL PART
3. 1 Solution linear equations with two unknowns
1. Solve the equation 407x - 2816y = 33 in whole numbers.
I will use the compiled algorithm.
1. Using the Euclidean algorithm, I will find the greatest common divisor of the numbers 407 and 2816:
2816 = 407 6 + 374;
407 = 374 1 + 33;
374 = 33 11 + 11;
Therefore (407.2816) = 11, with 33 divisible by 11.
2. Divide both sides of the original equation by 11, we get the equation 37x - 256y = 3, and (37, 256) = 1
3. Using the Euclidean algorithm, I will find a linear representation of the number 1 through the numbers 37 and 256.
256 = 37 6 + 34;
I will express 1 from the last equality, then successively going up the equalities I will express 3; 34 and substitute the resulting expressions into the expression for 1.
1 = 34 - 3 11 = 34 - (37 - 34 1) 11 = 34 12 - 37 11 = (256 - 37 6) 12 - 37 11 =
83 37 - 256 (- 12)
Thus, 37·(- 83) - 256·(- 12) = 1, therefore the pair of numbers x0 = - 83 and y0 = - 12 is a solution to the equation 37x - 256y = 3.
4. I will write down the general formula for solutions to the original equation where t is any integer.
Answer. (-83c+bt; -12c-at), t є Z.
Comment. It can be proven that if the pair (x1,y1) is an integer solution to the equation where, then all integer solutions to this equation are found by the formulas: x=x1+bty=y1-at
2. Solve the equation 14x - 33y=32 in whole numbers.
Solution: x = (32 + 33y) : 14
(14 [. ] 2+ 5)y + (14 [. ] 2 + 4) = 14 [. ] 2y + 5y + 14[. ] 2 + 4 = 14(2y + 2) + 5y + 4; 2y + 2 = p; p є Z
Search from 1 to 13
When y = 2; (5 [. ] 2 + 4): 14
Let me substitute y = 2 into the original equation
14x = 32 +33 [. ] 2
14x = 32 + 66 x = 98: 14 = 7
I will find all the integer solutions from the found quotient:
14(x - 7) + 98 - 33 (y -2) - 66 = 32
14(x - 7) - 33(y - 2)=0
14(x - 7) = 33(y - 2) -> 14(x - 7) : 33 -> (x - 7): 33 -> x = 33k + 7; k є Z
Let me substitute into the original equation:
14(33k + 7) - 33y = 32
14. 33k + 98 - 33y = 32 y = 14k + 2; x = 33k + 7, where k є Z. These formulas specify the general solution to the original equation.
Answer. (33k + 7; 14k + 2), k є Z.
3. Solve the equation x - 3y = 15 in whole numbers.
I will find GCD(1,3)=1
I will determine a particular solution: x=(15+3y):1 using the enumeration method, I find the value y=0 then x=(15+3 [. ] 0) =15
(15; 0) - private solution.
All other solutions are found using the formulas: x=3k + 15, k є Z y=1k+0=k, k є Z with k=0, I get a particular solution (15;0)
Answer: (3k+15; k), k є Z.
4. Solve the equation 7x - y = 3 in whole numbers.
I will find GCD(7, -1)=1
I will define a particular solution: x = (3+y):7
Using the brute force method, we find the value y є y = 4, x = 1
This means (1;4) is a particular solution.
I find all other solutions using the formulas: x = 1k + 1, k є Z y = 7k + 4, k є Z
Answer: (k+1;7k+4); k є Z.
5. Solve the equation 15x+11 y = 14 integers.
I will find GCD(15, -14)=1
I will define a particular solution: x = (14 - 11y):15
Using the brute force method, I find the value y є y = 4, x = -2
(-2;4) is a particular solution.
I find all other solutions using the formulas: x = -11k - 2, k є Z y =15k + 4, k є Z
Answer: (-11k-2; 15k+4); k є Z.
6. Solve the equation 3x - 2y = 12 integers.
I will find GCD(3; 2)=1
I will define a particular solution: x = (12+2y):3
Using the brute force method, I find the value y є y = 0, x = 4
(4;0) is a particular solution.
I find all other solutions using the formulas: x = 2k + 4, k є Z y = 3k, k є Z
Answer: (2k+4; 3k); k є Z.
7. Solve the equation xy = x + y in whole numbers.
I have xy - x - y + 1 = 1 or (x - 1)(y - 1) = 1
Therefore x - 1 = 1, y - 1 = 1, whence x = 2, y = 2 or x - 1 = - 1, y - 1 = - 1, whence x = 0, y = 0 other solutions in integers given the equation does not have.
Answer. 0;0;(2;2).
8. Solve the equation 60x - 77y = 1 in whole numbers.
Let me solve this equation for x: x = (77y + 1) / 60 = (60y + (17y +1)) / 60 = y + (17y + 1) / 60.
Let (17y + 1) / 60 = z, then y = (60z - 1) / 17 = 3z + (9z - 1) / 17. If we denote (9z - 1) / 17 by t, then z = (17t + 1) / 9 = 2t + (- t + 1) / 9. Finally, let (- t + 1) / 9 = n, then t = 1- 9n. Since I find only integer solutions to the equation, z, t, n must be integers.
Thus, z = 2 - 18n + 2 = 2 - 17n, and therefore y = 6 - 51n + 1 - 9n = 7 - 60n, x = 2 - 17n +7 - 60n = 9 - 77n. So, if x and y are integer solutions of a given equation, then there is an integer n such that x = 9 - 77n, y = 7 - 60n. Conversely, if y = 9 - 77n, x = 7 - 60n, then, obviously, x, y are integers. Check shows that they satisfy the original equation.
Answer. (9 - 77n; 7 - 60n)); n є Z.
9. Solve the equation 2x+11y =24 in whole numbers.
I will find GCD(2; 11)=1
I will define a particular solution: x = (24-11y):2
Using the brute force method, I find the value y є y = 0, x = 12
(12;0) is a particular solution.
I find all other solutions using the formulas: x = -11k + 12, k є Z y = 2k + 0=2k, k є Z
Answer:(-11k+12; 2k); k є Z.
10. Solve the equation 19x - 7y = 100 in whole numbers.
I will find GCD(19, -7)=1
I will define a particular solution: x = (100+7y):19
Using the brute force method, I find the value y є y = 2, x = 6
(6;2) is a particular solution.
I find all other solutions using the formulas: x = 7k + 6, k є Z y = 19k + 2, k є Z
Answer:(7k+6; 19k+2); kє Z.
11. Solve the equation 24x - 6y = 144 in whole numbers
I will find GCD(24, 6)=3.
The equation has no solutions because GCD(24, 6)!=1.
Answer. There are no solutions.
12. Solve the equation in whole numbers.
I transform the ratio of coefficients for unknowns.
First of all, I will highlight the whole part of the improper fraction;
I will replace the proper fraction with an equal fraction.
Then I'll get it.
I will do the same transformations with the improper fraction obtained in the denominator.
Now the original fraction will take the form:
Repeating the same reasoning for the fraction, I get.
Isolating the whole part of the improper fraction, I come to the final result:
I got an expression called a finite continued fraction or continued fraction. Having discarded the last link of this continued fraction - one fifth, I will transform the resulting new continued fraction into a simple one and subtract it from the original fraction.
I will reduce the resulting expression to a common denominator and discard it, then
From comparing the resulting equality with the equation it follows that, will be a solution to this equation and, according to the theorem, all its solutions will be contained in,.
Answer. (9+52t; 22+127t), t є Z.
The result obtained suggests that in the general case, in order to find a solution to the equation, it is necessary to expand the ratio of the coefficients of the unknowns into a continued fraction, discard its last link and carry out calculations similar to those carried out above.
13. Solve the equation 3xy + 2x + 3y = 0 in integers.
3xy + 2x + 3y = 3y + 2x + 3y + 2 - 2 = 3y(x + 1) + 2(x + 1) - 2 =
=(x + 1)(3y + 2) - 2,
(x + 1)(3y + 2) = 2,
3y + 2 = 1 or 3y + 1 = 2 or 3y + 1 = -1 or 3y + 1 = -2 x + 1 = 2, x + 1 =1, x + 1 = -2, x + 1 = -1 ; x = 2 or x = 0 or x = -3 or x = -2 y cent z, y = 0, y = -1, y cent z.
Answer: (0;0);(-3;-1).
14. Solve the equation y - x - xy = 2 in whole numbers.
Solution: y - xy - x + 1 = 3, (y + 1)(1 - x) = 3,
3 = 1·3 = 3·1 = (-1)·(-3) = (-3)·(-1).
y + 1 = 1 or y + 1 = 3 or y + 1 = -1 or y + 1 = -3
1 - x =3, 1 - x =1, 1 - x = -3, 1 - x = -1.
y = 0 or y = 2 or y = -2 or y = -4 x = -2, x = 0, x = 4, x = 2
Answer: (-2;0);(0;2);(2;-4);(4;-2).
15. Solve the equation y + 4x + 2xy = 0 in whole numbers.
Solution: y + 4x + 2xy + 2 - 2 = 0, (2x + 1)(2 + y) = 2,
2 = 1∙2 = 2∙1 = (-2)∙(-1) = (-1)∙(-2).
2x + 1= 1 or 2x + 1= 2 or 2x + 1= -1 or 2x + 1= -2
2 + y = 2, 2 + y = 1, 2 + y = -2, 2 + y = -1; y = 0 or y = -1 or y = -4 or y = -3 x = 0, x cent Z, x = -1, x cent Z.
Answer: (-1;-4);(0;0).
16. Solve the equation 5x + 10y = 21 in whole numbers.
5(x + 2y) = 21, since 21 != 5n, then there are no roots.
Answer. There are no roots.
17. Solve the equation 3x + 9y = 51 in natural numbers.
3(x + 3y) = 3∙17, x = 17 - 3y, y = 1, x = 14; y = 2, x = 11; y = 3, x = 8; y = 4, x = 5; y = 5, x = 2; y = 6, x = -1, -1cent N.
Answer:(2;5);(5;4);(8;3);(11;2; (14:1).
18. Solve the equation 7x+5y=232 in whole numbers.
I will solve this equation with respect to the unknown at which the smallest (modulo) coefficient is found, that is, in this case with respect to y: y = 232-7x5.
Let me substitute the numbers instead of x into this expression: 0;1;2;3;4. I get: x=0, y=2325=4625, x=1, y=232-75=45, x=2, y=232-145=43.6, x=3, y=232-215=42, 2, x=4, y=232-285=40.8
Answer. (1;45).
19. Solve the equation 3x + 4y + 5xy = 6 in whole numbers.
I have 3∙4 + 5∙6 = 42 = mn
Divisors 42: - +- (1, 2, 3, 6, 7, 14, 21, 42).
x = m - 45, y = n - 35 I find that for m = -1, -6, 14, -21 n = -42, -7, 3, -2 the solutions are: x = -1, -2, 2, -5 y = -9, -2, 0, -1.
So, this equation has 4 solutions in integers and none in natural numbers.
Answer. -1;-9;-2;-2;2;0;(-5;-1).
20. Solve the equation 8x+65y=81 in natural numbers.
81⋮GCD(8;65)=>
8x=81-65y x=81-65y8=16+65-65y8=2+65(1-y)8.
Let 1-y8=t, t Є Z. x=2+65t>0y=1-8t>0
65t>-2-8t>-1 t>-265 t t=0.
At t=0 x=2y=1
Answer. (2;1).
21. Find integer non-negative solutions to the equation 3x+7y=250.
250⋮GCD(3;7) =>the equation can be solved in integers.
x=250-7y3=243+7-7y3=81+7(1-y)3.
Let 1-y3=t, t Є Z.
x=81+7t>=0y=1-3t>=0
7t>=-81-3t>=-1 t>=-817t=-1147t t=-11;-10;. ;0.
x=81+7tу=1-3t t=-11 x=4y=34 t=-10 x=11y=31 t=-9 x=18y=28 t=-8 x=25y=25 t=-7 x =32y=22 t=-6 x=39y=19 t=-5 x=46y=16 t=-4 x=53y=13 t=-3 x=60y=10 t=-2 x=67y=7 t =-1 x=74y=4 t=0 x=81y=1
Answer. 11;31;18;28;25;25;32;22;39;19;46;16;53;13;60;10;67;7;74;4;81;1.
22. Solve the equation xy+x+y3=1988 in integers.
Let's multiply both sides of the equation by 3. We get:
3x+3xy+y=5964
3x+3xy+y+1=5965
(3х+1)+(3х+у)=5965
(3x+1) + y(3x+1)=5965
(3x+1)(y+1)=5965
5965=1∙5965 or 5965=5965∙1 or 5965=-1∙(-5965) or 5965=-5965∙(-1) or 5965=5∙1193 or 5965=1193∙1 or 5965=-5∙( -1193) or 5965=-1193∙(-5)
1) 3x+1=1y+1=5965 2) 3x+1=5965y+1=1 x=0y=5964 x=1988y=0
3) 3x+1=5y+1=1193 4) 3x+1=1193y+1=5 solutions in integers no solutions in integers no
5) 3x+1=-1y+1=-5965 6) 3x+1=-5965y+1=-1 no solutions in integers no solutions in integers
7) 3x+1=-5y+1=-1193 8) 3x+1=-1193y+1=-5 x=-2y=1194 x=-398y=-6
Answer. 0;5964;1988;0;-2;-1194;(-398;-6).
3. 2 SOLVING PROBLEMS
There are several types of problems, most often these are problems of an Olympiad nature, which boil down to solving Diophantine equations. For example: a) Tasks on exchanging a sum of money of a certain denomination.
b) Problems involving transfusion and dividing objects.
1. We bought 390 colored pencils in boxes of 7 and 12 pencils. How many of these and other boxes did you buy?
I will designate: x boxes of 7 pencils, y boxes of 12 pencils.
Let me create an equation: 7x + 12y = 390
I will find GCD(7, 12)=1
I will define a particular solution: x = (390 - 12y):7
Using the brute force method, I find the value y є y = 1, x = 54
(54;1) is a particular solution.
I find all other solutions using the formulas: x = -12k + 54, k є Z y = 7k + 1, k є Z
I found many solutions to the equation. Taking into account the conditions of the problem, I will determine the possible number of both boxes.
Answer. You can buy: 54 boxes of 7 pencils and 1 box of 12 pencils, or 42 boxes of 7 pencils and 8 boxes of 12 pencils, or 30 boxes of 7 pencils and 15 boxes of 12 pencils, or 28 boxes of 7 pencils and 22 boxes of 12 pencils , or 6 boxes of 7 pencils and 29 boxes of 12 pencils.
2. One leg of a right triangle is 7 cm larger than the other, and the perimeter of the triangle is 30 cm. Find all the sides of the triangle.
I will designate: x cm - one leg, (x+7) cm - the other leg, y cm - hypotenuse
I will compose and solve the Diophantine equation: x+(x+7)+y=30
I will find GCD(2; 1)=1
I will define a particular solution: x = (23 - y):2
Using the brute force method, I find the value y =1 y = 1, x = 11
(11;1) is a particular solution.
I find all other solutions to the equation using the formulas: x = -k + 11, k є Z y = 2k + 1, k є Z k
Considering that any side of the triangle less than the amount two other sides, we come to the conclusion that there are three triangles with sides 7, 9 and 14; 6, 11 and 13; 5, 13 and 12. According to the conditions of the problem, given right triangle. This is a triangle with sides 5, 13 and 12 (the Pythagorean theorem holds).
Answer: One leg is 5 cm, the other is 12 cm, the hypotenuse is 13 cm.
3. Several children were picking apples. Each boy collected 21 kg, and the girl collected 15 kg. In total they collected 174 kg. How many boys and how many girls picked apples?
Let there be x boys and y girls, with x and y being natural numbers. Let me create an equation:
I solve by selection method: x
6 Only at x = 4 does the second unknown receive a positive integer value (y = 6). For any other value of x, y will be either a fraction or negative. Therefore, the problem has one the only solution.
Answer. 4 boys and 6 girls.
4. Is it possible to create a set of pencils worth 3 rubles and pens worth 6 rubles worth 20 rubles?
Let the number of pencils in the set be x and the number of pens be y.
Let me create an equation:
For any integers x and y, the left side of the equation must be divisible by 3; the right-hand side is not divisible by 3. This means that there are no integers x and y that would satisfy our equation. This equation cannot be solved in integers. It is impossible to create such a set.
Answer. There are no solutions.
5. Find a natural number that, when divided by 3, leaves a remainder of 2, and when divided by 5, leaves a remainder of 3.
I will denote the required number by x. If I denote the quotient of x by 3 by y, and the quotient of division by 5 by z, then I get: x=3y+2x=5z+3
According to the meaning of the problem, x, y and z must be natural numbers. This means that we need to solve an indefinite system of equations in integers.
For any integer y and z, x will also be an integer. I subtract the first from the second equation and get:
5z - 3y + 1 = 0.
Having found all positive integers y and z, I will immediately get all integers positive values x.
From this equation I find:
One solution is obvious: for z = 1 we get y = 2, and x and y are integers. The solution x = 8 corresponds to them.
I'll find other solutions. To do this, I will introduce an auxiliary unknown u, setting z = 1 + u. I will receive:
5(1 + u) - 3y + 1 = 0, i.e. 5u = 3y - 6 or 5u = 3(y - 2).
The right-hand side of the last equation is divisible by 3 for any integer y. This means that the left-hand side must also be divisible by 3. But the number 5 is coprime to the number 3; therefore u must be divisible by 3, i.e., have the form 3n, where n is an integer. In this case y will equal
15n/3 + 2 = 5n + 2, i.e., also an integer. So, z = 1 + u = 1 + 3n, whence x = 5z + 3 = 8 + 15n.
The result is not one, but an infinite set of values for x: x = 8 + 15n, where n is an integer (positive or zero):
Answer. x=8+15n; n є 0;1;2;.
6. Subjects brought 300 as gifts to the Shah precious stones: small boxes contain 15 pieces each and large boxes contain 40 pieces. How many of these and other boxes were there, if it is known that there were fewer small ones than large ones?
Let me denote by x the number of small boxes, and by y the number of large ones.
15x+40y=300. I'll cut it by 5.
3x+8y=60 x=60-8y3 x=60-6y-2y3
X=20-2y-2y3
For the value of a fraction to be an integer, 2y must be a multiple of 3, i.e. 2y = 3c.
I will express the variable y and select the whole part:
Z must be a multiple of 2, i.e. z=2u.
Let me express the variables x and y in terms of u:
X=20-2y-2y3
Х=20-2∙3u-2∙3u3
I will compose and solve a system of inequalities:
I will write down the entire solutions: 1; 2. Now I will find the values of x and y for u=1; 2.
1) x1=20-8∙1=20-8=12 y1=3∙1=3
2) x2=20-8∙2=20-16=4 y2=3∙2=6
Answer. 4 small boxes; 6 large boxes.
7. Two Ural 5557 cars were given, the cars were sent on a flight Krasnoturinsk - Perm - Krasnoturinsk. In total, 4 tons of diesel fuel and 2 drivers were needed to complete this flight. It is necessary to determine transport costs, namely the cost of 1 ton of diesel fuel and wages for drivers performing this flight, if it is known that a total of 76,000 rubles were spent.
Let x rubles be the cost of 1 ton of diesel fuel, and let x rubles be the wages of drivers. Then (4x + 2y) rubles were spent on the flight. And according to the conditions of the problem, 76,000 rubles were spent.
I get the equation:
To solve this equation, the brute-force method will be a labor-intensive process. So I'll use the > method.
I’ll express the variable y through x: , select the whole part, and get: (1).
For the value of a fraction to be an integer, 2x must be a multiple of 4. That is, 2x = 4z, where z is an integer. From here:
I’ll substitute the value of x into expression (1):
Since x, y 0, then 19000 z 0, therefore, giving z integer values from 0 to 19000, I get the following values of x and y: z
From real data on transport costs, it is known that 1 ton of diesel fuel (x) costs 18,000 rubles. , and the payment for drivers performing flight (y) is 10,000 rubles. (data taken approximately). From the table we find that the x value equal to 18000 and the y value equal to 10000 correspond to a z value equal to 9000, indeed: ;.
8. In how many ways can you collect the amount of 27 rubles? , having quite a lot of two-ruble and five-ruble coins?
Let me denote: x two-ruble coins and y five-ruble coins
I will create an equation, taking into account the condition of the problem 2x + 5y = 27.
I will find GCD(2;5)=1
I will define a particular solution: x = (27-5y):2
Using the brute force method, I find the value y є y = 1, x = 11
(11;1) is a particular solution.
All other solutions are found using the formulas: x = -5k + 11, k є Z y = 2k + 1, k є Z
This equation has many solutions. Let's find all the ways in which you can collect the amount of 27 rubles with the offered coins. k
Answer. There are three ways in which you can collect this amount if you have a lot of two-ruble and five-ruble coins.
9. Let's say octopuses and starfish live in an aquarium. Octopuses have 8 legs, and starfish have 5. There are 39 limbs in total. How many animals are there in the aquarium?
Let x be the number of starfish, y the number of octopuses. Then all octopuses have 8 legs, and all stars have 5 legs.
Let me create an equation: 5x + 8y = 39.
I note that the number of animals cannot be expressed as a whole or negative numbers. Therefore, if x is a non-negative integer, then y = (39 - 5x)/8 must also be an integer and non-negative, and, therefore, it is necessary that the expression 39 - 5x be divisible by 8 without a remainder. A simple search of options shows that this possible only when x = 3, then y = 3.
Answer: (3; 3).
10. On furniture factory They make stools with three and four legs. The master made 18 legs. How many stools can be made so that all the legs can be used?
Let x be the number of three-legged stools and y the number of four-legged ones. Then, 3x + 4y = 18.
I have, 4y =18 - 3x; y = 3(6 - x):4.
I get: x = 2; y = 3 or x = 6; y = 0.
There are no other solutions, since x 6.
Answer. 2;3;(6;0).
11. Is it possible to accommodate 718 people in 4- and 8-berth cabins, so that there are no empty seats in the cabins?
Let the 4-bed cabins be x, and the 8-bed cabins y, then:
2(x + 2y) = 309
Answer. It is forbidden.
12. Prove that on the line 124x + 216y = 515 there is not a single point with integer coordinates.
GCD(124,216) = 4, 515 != 4n, which means there are no integer solutions.
Answer. There are no solutions.
13. The cost of the goods is 23 rubles, the buyer has only 2 ruble coins, and the cashier has 5 ruble coins. Is it possible to make a purchase without first exchanging money?
Let x be the number of 2 ruble coins, y the number of 5 ruble coins, then 2x - 5y = 23, where x,y є N.
I get: 2x = 23 + 5y, from where x =23 + 5y2 =11 + 2y + (1 + y)2 x will be an integer if 1 + y2 is an integer.
1 + y2 = t, where t Euro Z, then y = 2t - 1.
x = 11 + 2y + 1 + y2 = 11 + 4t - 2 + 1 + 2t-12 = 5t + 9.
T. o. x = 5t + 9, and y = 2t - 1, where t є z.
The problem has many integer solutions. The simplest of them is for t = 1, x =14, y = 1, i.e. the buyer will give fourteen 2-ruble coins and receive one 5-ruble coin in change.
Answer. Can.
14. During an audit of the store’s trade books, one of the entries turned out to be covered in ink and looked like this:
> It was impossible to make out the number of meters sold, but there was no doubt that the number was not a fraction; in the proceeds only three could be distinguished last digits, and also establish that there were three other numbers in front of them. Is it possible to restore a record using this data?
Let the number of meters be x, then the cost of the goods in kopecks is 4936x. We denote the three filled in digits in total as y, this is the number of thousands of kopecks, and the entire amount in kopecks will be expressed as follows (1000y + 728).
I get the equation 4936x = 1000y + 728, I divide it by 8.
617x - 125y = 91, where x,y є z, x,y
125y = 617x - 91 y = 5x - 1 +34 - 8x125 = 5x - 1 + 2 17 - 4x125 =
5x - 1 + 2t, where t = 17 - 4x125, t Euro Z.
From the equation t = (17 - 4x)/125 I get x = 4 - 31t + 1 - t4 =
4 - 31t + t1, where t1 = 1 - t4, hence t = 1 - 4t1, a x = 125t1 - 27, y = 617t1 - 134.
By condition I know that 100
100 = 234/617 and t1
This means that 98 meters were sold for the amount of 4837.28 rubles. The recording has been restored.
Answer. 98 meters released.
15. It is required to buy 40 pieces for one ruble postage stamps- kopecks, 4-kopecks and 12 kopecks. How many stamps of each denomination can you buy?
You can make two equations: x + 4y + 12z = 100 and x + y + z = 40, where x is the number of penny marks, y is the number of 4-kopeck marks, z is the number of 12-kopeck marks. I subtract the second from the first equation and get:
3y + 11z = 60, y = 60 - 11z3 = 20 - 11· z3.
Let z3 = t, z = 3t, where t Euro Z. Then I get if x + y + z = 40 and z = 3t, and y = 20 - 11t, x = 20 + 8t.
Since x >= 0, y >= 0, z >= 0, then 0
Then, accordingly, I get: t = 0, x = 20, y = 20, z = 0; t = 1, x = 28, y = 9, z = 3.
So, the purchase of stamps can be made in only two ways, and if the condition is that at least one stamp of each denomination be purchased, then only in one way.
Answer. 28 marks of 1 kopeck, 9 marks of 4 kopecks and 3 marks of 12 kopecks.
16. A student was given a task of 20 problems. For each correctly solved question, he receives 8 points; for each unsolved question, 5 points are deducted from him. For a task that he did not undertake - 0 points. The student scored 13 points in total. How many problems did he undertake to solve?
Let the correctly solved problems be x, the incorrectly solved problems be y, and the not considered problems be z.
Then x + y + z = 20, and 8x - 5y = 13.
y = 8x - 135= x - 2 +3(x - 1)5 = x - 2 + 3t, where t = x - 15, and x = 5t + 1.
By condition x + y
Answer: the student took on 13 problems, solved 6, and failed 7.
17. Ivanushka the Fool fights with the Serpent Gorynych, who has 2001 heads. Swinging his sword to the left, Ivan cuts off 10 heads, and in return 16 grow. Swinging his sword to the right, he cuts off 15, and 6 grow. If all the heads are cut off, no new ones grow. You can swing in any order, but if there are less than 15 goals, then only to the left, and if there are less than 10, then not at all. Can Ivanushka the Fool defeat the Serpent Gorynych?
Let me rephrase the problem: is it possible to cut down 1986 heads? Then Ivan will cut down the remaining 15 with one blow to the right and no new ones will grow.
Let x be the number of strokes to the right, and y the number of strokes to the left, then 1986 - 9x + 6y = 0.
I divide the whole equation by 6, I get
3x - 2y = 662.
y = 3x - 6622 = x - 331 + x2.
Let x2 = t, then x = 2t, and y = 3t - 331.
Since x >= 0, y >= 0, then t >= 111, hence t = 111, x = 222, y = 2.
I get: by hitting 220 times to the right, Ivan cuts off 1980 heads and the Serpent has 21 heads left; then 2 hits to the left and the Snake grows 12 heads, making a total of 33; the next 2 blows to the right deprive the Snake of 18 heads and Ivan cuts off the remaining 15 with the last blow to the right and no new heads grow.
Answer: 220 strikes to the right, 2 strikes to the left and 3 more strikes to the right.
18. The sides of a dice are numbered - 1, 2, 3, 4, 5, 6. From 5 such cubes, they built a tower and counted the sum of points on all visible faces, after removing the top cube, the sum decreased by 19, which number turned out to be the top edge of the top cube?
The sum of points of one cube is 21.
Let x be the number of points on the bottom face of the top cube, and y the number of points on the top face of the next cube. When you remove the top cube, the points of 5 faces of the top cube disappear, the sum of the points of which is (21 - x), and the face on which the points appears, which means that the sum of the points has decreased by (21 - x) - y, and according to the condition it is 19, hence :
(21 - x) - y = 19, x + y = 2.
Hence y = 2 - x, and by condition 1
19. Someone bought 30 birds for 30 coins of the same denomination. For every 3 sparrows you pay 1 coin, for 2 bullfinches - 1 coin, for 1 dove - 2 coins. How many birds of each type were there?
Let there be x sparrows, y bullfinches, and z pigeons. Then, according to the condition x + y + z = 30 and 13x + 12y + 2z = 30.
I get x + y + z = 30 and 2x + 3y + 12z = 180, or y + 10z = 120, y = 120 - 10z, where by condition x
Hence the following options (0;20;10); (9;10;11); (18;0;12).
Answer: sparrows - 0, bullfinches - 20, pigeons - 10; sparrows - 9, bullfinches - 10, pigeons - 11; sparrows - 18, bullfinches - 0, pigeons - 12.
20. Find all two-digit numbers, each of which, when reduced by 2, is equal to five times the product of its digits.
Let xy be the required two-digit numbers.
For the equation xy - 2 = 5xy, or (10x + y) - 5xy = 2 S = 0 and I will find all natural solutions from the set (x; 2).
Since x is the first digit of two-digit numbers, it can only take 9 values.
That. , the required numbers will be: 12, 22, 32,. , 92.
Answer. 12; 22, 32; 42; 52; 62; 72; 82; 92.
21. A piece of wire 102 cm long needs to be cut into pieces 15 cm and 12 cm long so that all the wire is used. How to do this?
Let x be the number of pieces of wire 15 cm long, y be the number of pieces of wire 12 cm long. Let’s create an equation:
15x+12y=102 /:3
4x+3y=34 x=34-4y5=6+4-4y5=6+4(1-y)5.
Let 1-y5=t x=6+4t>0y=1-5t>0=> 4t>-6-5t>-1 => t>-1.5t t=0;-1.
If t=0, then x=6y=1
If t=-1, then x=2y=6
Answer. The problem has two solutions:
1) 102=15∙6+12∙1; 2) 102=15∙2+12∙6.
22. Petya in 1987 was as old as the sum of the digits of the year of his birth. What year was he born?
Let Petya be born in 1919. Then in 1987 he was 1987-19xy, or (1+9+x+y) years old. We have the equation:
87-(10x+y)=10+x+y
77-11x=2y y=77-11x2=38-11x-12.
Considering that x and y are digits of the decimal number system, we find by selection: x=3, y=1.
Answer. Petya was born in 1970.
23. Someone buys an item worth 19 rubles in a store. He has only 15-three-ruble notes, while the cashier has only 20-five-ruble notes. Can I pay and how?
The problem comes down to solving the Diophantine equation in positive integers: 3x - 5y = 19, where x
Due to the fact that x>0 and y > 0 and taking into account the conditions of the problem, it is easy to establish that 0
This implies 2 possible values: x
Answer. 1) 19=3∙8-1∙5 2) 19=3∙13-4∙5.
24. Is it possible to weigh 28 g of a certain substance on a cup scale, having only 4 weights weighing 3 g and 7 weights weighing 5 g?
To do this you need to solve the equation:
x = 9 - 2(3y1 - 1) + y1 = 11-5y1.
So x = 11 - 5 y1 y = 3 y1 - 1.
It follows from the conditions of the problem that y1 cannot be given negative values. Next should be y1
Answer. 1 weight in 3 g and 5 weights in 5 g.
25. The buyer purchased in the store for 21 rubles. goods. But he only has banknotes of 5 ruble denominations, while the cashier has 3 ruble ones. You want to know if you can pay the cashier if you have money and how exactly?
Let x be the number 5 - rubles, y - 3 - rubles.
By condition, x > 0, y > 0, that means.
Also, t is even, otherwise neither x nor y will be integer.
At t = 4, 6, 8,. we have: t
Answer. 6;3;8;8;12;13;15;18;18;23;21;28;24;33;27;38;(30;43).
26. There are 110 sheets of paper. It is required to sew notebooks of 8 sheets and 10 sheets each. How many do you need to sew?
Let x be the number of 8 sheet notebooks, y the number 10 sheet notebooks.
So t = 0 or t = - 1
Answer. 5;7;(10;3).
27. Many old ways Guessing numbers and dates of birth is based on solving Diophantine equations. For example, to guess the date of birth (month and day) of your interlocutor, it is enough to ask him for the sum obtained from adding two products: the date number (x) by 12 and the month number (y) by 31.
Let the sum of the products about which we're talking about, is equal to 330. Find the date of birth.
Let's solve the indeterminate equation: y = 2y1 + y2 = 2(2y2 + y3) + y2 = 5y2 + 2y3 = 5(2y3 - 6) + 2y3 = 12y3 - 30 x = 27 - 3(12y3 - 30) + 2y2 + y3 = 27 - 36y3 + 90 + 2(2y3 - 6) + y3 =
27 - 36y3 + 90 + 5y3 - 12 = 105 - 31y3 x = 12y3 - 30, y = 105 - 31y3
So, date of birth: 12th day of the 6th month.
28. Is it possible to collect the amount of 51 rubles with two-ruble and five-ruble coins? If possible, how many ways are there?
Let there be x two-ruble coins, and five-ruble coins.
Let 1+y2=z, then
=> z = 1, 2, 3, 4, 5
Answer: 5 ways.
29. Is it possible to put two hundred eggs in boxes of 10 and 12 pieces? If possible, find all such ways.
Let there be x boxes of 10 pieces each and let the boxes have 12 pieces each. Let me create an equation: z = 1, 2, 3
Answer: 14;5;8;10;(2;15)
30. Imagine the number 257 as the sum of two natural terms: a) one of which is a multiple of 3, and the other is a multiple of 4; b) one of which is a multiple of 5, and the other is a multiple of 8.
Answer: 1) 249 and 8; 2) 225 and 32.
In problems involving indefinite equations, I encountered a wide variety of cases: the problem may be completely unsolvable (Problem 4), may have an infinite number of solutions (Problem 2), may have several definite solutions; in particular, it can have one unique solution (Problem 1).
CONCLUSION
The goal that I set for myself has been achieved. Working on the project aroused interest and captivated me. This work required from me not only certain mathematical knowledge and perseverance, but also gave me the opportunity to feel the great joy of independent discovery.
Diophantine equations are found in Olympiad tasks, so they develop logical thinking, increase the level of mathematical culture, instill skills of independent research work in mathematics.
When solving equations and problems that reduce to Diophantine equations, the properties are applied prime numbers, polynomial factorization method, brute force method, descent method and Euclidean algorithm. In my opinion, the descent method is the most difficult. But the brute force method turned out to be prettier for me.
I solved 54 problems in my work.
This work has contributed to a deeper understanding school curriculum and broadening your horizons.
This material will be useful to students interested in mathematics. It can be used in some lessons and extracurricular activities.
Problem 62:
Solve the equation 3x + 5y = 7 in whole numbers. Solution:
Let's first find some specific solution (this idea, by the way, often helps in solving other problems). Since 3 2 + 5 (- 1) = 1, then 3 14 + 5 (- 7) = 7 and, therefore, x 0 = 14, y 0 = - 7 is the solution to our equation (one of many, no more!). So,
Subtract one equation from the other, denote x - x 0 and y - y 0 by a and b, and get 3a + 5b = 0. From here we see that b is divisible by 3, and a is divisible by 5. Let's put a = 5k, then b = - 3k - here k, obviously, can be any integer. So we get a set of solutions:
Where k can be any integer. Of course, there are no other solutions. Problem 63:
Find all integer solutions to the equation 3x - 12y = 7. Solution:
This equation has no integer solutions. The left side is divisible by 3, while the right side is not divisible by 3.
Problem 64:Solve the equation 1990x - 173y = 11. Solution:
The numbers involved in the formulation are so large that it is impossible to find a specific solution by selecting it. However, it will help us to know that the numbers 1990 and 173 are coprime (check this).
Lemma. Their gcd equal to 1 can be represented as 1990m - 173n, where m and n are some integers.
The proof of this lemma follows from the fact that all numbers that are obtained in the process of the Euclidean algorithm are representable in the specified form.
Specifically, in this case, using the Euclidean algorithm, you can get m = 2, n = 23. So, using such a powerful weapon as the Euclidean algorithm, we get a specific solution to the auxiliary equation 1990m - 173n = 1: pair (2, 23) . Therefore, x 0 = 22, y 0 = 253 is the solution to the equation 1990x - 173y = 11. Next we find that
K is any integer. Problem 65:
Find all integer solutions to the equation 21x + 48y = 6. Solution:
x = 16k - 2, y = - 7k + 1; k - any integer.
Problem 66:Solve the equation 2x + 3y + 5z = 11 in whole numbers. Solution:
x = 5p + 3q - 11, y = 11 - 5p - 2q, z = p; p, q - any integers.
Problem 67:The chip stands on one of the fields of an endless checkered strip of paper in both directions. It can be shifted m fields to the right or n fields to the left. For what m and n can she move to the cell adjacent to the right? In what minimum number of moves can she do this? Solution:
For relatively prime m and n.
Problem 68:(2x + y)(5x + 3y) = 7. Solution:
(- 4,9), (14, - 21), (4, - 9), (- 14,21).
Problem 69:xy = x + y + 3. Solution:
Since xy - x - y = 3, then (x - 1)(y - 1) = 4. All that remains is to sort out the possible decompositions of the number 4 into the product of two integer factors. Answer: (x = 5,y = 2), (2.5), (0, - 3), (- 3.0), (3.3), (- 1, - 1).
Problem 70:x² = 14 + y². Solution:
There are no solutions in whole numbers.
Problem 71:x² + y² = x + y + 2. Solution:
(2,0), (2,1), (- 1,0), (- 1,1), (0,2), (1,2), (0, - 1), (1, - 1).
This is how problem 69 is solved. Since xy - x - y = 3, then (x - 1)(y - 1) = 4. All that remains is to sort out the possible decompositions of the number 4 into the product of two integer factors. Answer: (x = 5,y = 2), (2.5), (0, - 3), (- 3.0), (3.3), (- 1, - 1).
Problem 72:x² + y² = 4z - 1.
In fact, let's see what remainders can give exact squares modulo 4 (the choice of modulus 4 is suggested to us by the very appearance of the right-hand side of the equation). A short search shows that these are remainders 0 and 1. Since the sum of two remainders of this type cannot give a remainder - 1, we find that this equation has no solutions.
Problem 73:x² - 7y = 10. Solution:
There are no solutions in whole numbers (Module 7).
Problem 74:x³ + 21y² + 5 = 0. Solution:
Since x³ can only be comparable modulo 7 with 0, 1 and - 1, then the expression x³ + 21y² + 5 is comparable (mod %)%7 with 5, 6 or 4, and therefore cannot be equal to zero.
Problem 75:15x² - 7y² = 9. Solution:
There are no solutions in whole numbers (Module 5).
Problem 76:x² + y² + z² = 8t - 1. Solution:
There are no solutions in whole numbers (Module 8).
Problem 77:3 m + 7 = 2 n. Solution:
Modulo 3, the left-hand side is comparable to 1, and from this we conclude that n is even, i.e. n = 2k. The equation is transformed to the form 3 m + 7 = 4 k. Now module 4 is included in the game. 4 k - 7 = 1 (mod %)%4, and we see that m is even, i.e. m = 2p. So we have the equation 3 2p + 7 = 2 2k. Let's transform the equation: 7 = 2 2k - 3 2p = (2 k - 3 p )(2 k + 3 p ). From here 2 k + 3 p = 7, 2 k - 3 p = 1, and we get the only solution k = 2, p = 1, i.e. m = 2, n = 4.
Problem 78:3 2 m + 1 = n². Solution:
It is immediately clear that n is not divisible by 3 and, therefore, n = 3k + 1 or n = 3k + 2. Let us examine both cases.
a) n = 3k + 2, 3 2 m + 1 = 9k² + 12k + 4. Reducing, we get 2 m = 3k² + 4k + 1 = (3k + 1)(k + 1). Therefore, both k + 1 and 3k + 1 are powers of two. It can be seen that both k = 0 and k = 1 are suitable, and we get solutions n = 2, m = 1 and n = 5, m = 3. But for k ≥ 2 4(k + 1) > 3k + 1 > 2 (k + 1) and therefore k + 1 and 3k + 1 cannot simultaneously be powers of two.
b) n = 3k + 1. Analyzing this case in a similar way, we obtain another solution n = 7, m = 4.
Problem 79:1/a + 1/b + 1/c = 1. Solution:
a = b = c = 3; a,b,c = 1,2,3 or 2,4,4; one of the numbers is 1, and the sum of the other two is 0, for example, a = 1, b = - c = 13.
Problem 80:x² - y² = 1988. Solution:
x = ± 498, y = ± 496 or x = ± 78, y = ± 64, and the signs are chosen independently.
Problem 81:Prove that the equation 1/x - 1/y = 1/n has a unique solution in natural numbers if and only if n is a prime number. Solution:
If n = pq (p, q > 1), then 1/n = 1/(n - 1) - 1/n(n - 1) and 1/n = 1/p(q - 1) - 1/pq (q - 1). If n is prime, then n(y - x) = xy, which means xy is divisible by n, i.e. x or y is divided by n. It is clear that it is y that is divisible by n: y = kn. Then x = kn/(n + 1), whence k = n - 1, i.e. there is exactly one representation 1/n = 1/(n - 1) - 1/n(n - 1).
Problem 82:Solve the equation in whole numbers: x³ + 3 = 4y(y + 1).
Problem 83:Solve the equation in whole numbers: x² + y² = z².
Problem 84:Solve the equation in whole numbers: x² - 5y² = 1.
