In the 7th grade mathematics course, they first meet with equations with two variables, but they are studied only in the context of systems of equations with two unknowns. That is why a number of problems fall out of sight, in which certain conditions are introduced on the coefficients of the equation that limit them. In addition, methods for solving problems like “Solve an equation in natural or integer numbers” are also ignored, although in USE materials and on entrance exams Problems of this kind are becoming more and more common.

Which equation will be called an equation with two variables?

So, for example, the equations 5x + 2y = 10, x 2 + y 2 = 20, or xy = 12 are two-variable equations.

Consider the equation 2x - y = 1. It turns into a true equality at x = 2 and y = 3, so this pair of variable values ​​is the solution to the equation under consideration.

Thus, the solution of any equation with two variables is the set of ordered pairs (x; y), the values ​​of the variables that this equation turns into a true numerical equality.

An equation with two unknowns can:

a) have one solution. For example, the equation x 2 + 5y 2 = 0 has only decision (0; 0);

b) have multiple solutions. For example, (5 -|x|) 2 + (|y| – 2) 2 = 0 has 4 solutions: (5; 2), (-5; 2), (5; -2), (-5; - 2);

in) have no solutions. For example, the equation x 2 + y 2 + 1 = 0 has no solutions;

G) have infinitely many solutions. For example, x + y = 3. The solutions to this equation will be numbers whose sum is 3. The set of solutions to this equation can be written as (k; 3 - k), where k is any real number.

The main methods for solving equations with two variables are methods based on factoring expressions, highlighting the full square, using properties quadratic equation, limited expressions, evaluation methods. The equation, as a rule, is transformed into a form from which a system for finding unknowns can be obtained.

Factorization

Example 1

Solve the equation: xy - 2 = 2x - y.

Solution.

We group the terms for the purpose of factoring:

(xy + y) - (2x + 2) = 0. Take out the common factor from each bracket:

y(x + 1) – 2(x + 1) = 0;

(x + 1)(y - 2) = 0. We have:

y = 2, x is any real number or x = -1, y is any real number.

In this way, the answer is all pairs of the form (x; 2), x € R and (-1; y), y € R.

Equality to zero of non-negative numbers

Example 2

Solve the equation: 9x 2 + 4y 2 + 13 = 12(x + y).

Solution.

Grouping:

(9x 2 - 12x + 4) + (4y 2 - 12y + 9) = 0. Now each parenthesis can be collapsed using the square difference formula.

(3x - 2) 2 + (2y - 3) 2 = 0.

The sum of two non-negative expressions is zero only if 3x - 2 = 0 and 2y - 3 = 0.

So x = 2/3 and y = 3/2.

Answer: (2/3; 3/2).

Evaluation method

Example 3

Solve the equation: (x 2 + 2x + 2) (y 2 - 4y + 6) = 2.

Solution.

In each bracket, select the full square:

((x + 1) 2 + 1)((y – 2) 2 + 2) = 2. Estimate the meaning of the expressions in brackets.

(x + 1) 2 + 1 ≥ 1 and (y - 2) 2 + 2 ≥ 2, then the left side of the equation is always at least 2. Equality is possible if:

(x + 1) 2 + 1 = 1 and (y - 2) 2 + 2 = 2, so x = -1, y = 2.

Answer: (-1; 2).

Let's get acquainted with another method for solving equations with two variables of the second degree. This method is that the equation is considered as square with respect to some variable.

Example 4

Solve the equation: x 2 - 6x + y - 4√y + 13 = 0.

Solution.

Let's solve the equation as a quadratic one with respect to x. Let's find the discriminant:

D = 36 - 4(y - 4√y + 13) = -4y + 16√y - 16 = -4(√y - 2) 2 . The equation will have a solution only when D = 0, i.e., if y = 4. We substitute the value of y into the original equation and find that x = 3.

Answer: (3; 4).

Often in equations with two unknowns indicate restrictions on variables.

Example 5

Solve the equation in integers: x 2 + 5y 2 = 20x + 2.

Solution.

Let's rewrite the equation in the form x 2 = -5y 2 + 20x + 2. The right side of the resulting equation, when divided by 5, gives a remainder of 2. Therefore, x 2 is not divisible by 5. But the square of a number that is not divisible by 5 gives a remainder of 1 or 4. Thus equality is impossible and there are no solutions.

Answer: no roots.

Example 6

Solve the equation: (x 2 - 4|x| + 5) (y 2 + 6y + 12) = 3.

Solution.

Let's select the full squares in each bracket:

((|x| – 2) 2 + 1)((y + 3) 2 + 3) = 3. The left side of the equation is always greater than or equal to 3. Equality is possible if |x| – 2 = 0 and y + 3 = 0. Thus, x = ± 2, y = -3.

Answer: (2; -3) and (-2; -3).

Example 7

For each pair of negative integers (x; y) satisfying the equation
x 2 - 2xy + 2y 2 + 4y = 33, calculate the sum (x + y). Answer the smallest amount.

Solution.

Select full squares:

(x 2 - 2xy + y 2) + (y 2 + 4y + 4) = 37;

(x - y) 2 + (y + 2) 2 = 37. Since x and y are integers, their squares are also integers. The sum of the squares of two integers, equal to 37, we get if we add 1 + 36. Therefore:

(x - y) 2 = 36 and (y + 2) 2 = 1

(x - y) 2 = 1 and (y + 2) 2 = 36.

Solving these systems and taking into account that x and y are negative, we find solutions: (-7; -1), (-9; -3), (-7; -8), (-9; -8).

Answer: -17.

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Object of study.

The research concerns one of the most interesting branches of number theory - the solution of equations in integers.

Subject of study.

The solution in integers of algebraic equations with integer coefficients in more than one unknown is one of the most difficult and ancient mathematical problems and is not presented in sufficient depth in school course mathematics. In my work, I will present a fairly complete analysis of equations in integers, a classification of these equations according to the methods for solving them, a description of the algorithms for solving them, as well as practical examples of the application of each method for solving 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:

Faced with equations in integers in the Olympiad tasks, I assumed that the difficulties in solving them are due to the fact that not all the ways to solve them are known to me.

Relevance:

When solving approximate variants of USE tasks, I noticed that there are often tasks for solving equations of the first and second degree in integers. In addition, the Olympiad tasks various levels also contain equations in integers or problems that are solved using the skills to solve equations in integers. The importance of knowing how to solve equations in integers determines the relevance of my research.

Research methods

Theoretical analysis and generalization of information scientific literature about equations in integers.

Classification of equations in integers according to the methods of their solution.

Analysis and generalization of methods for solving equations in integers.

Research results

The paper describes methods for solving equations, considers the theoretical material of Fermat's theorem, the Pythagorean theorem, Euclid's algorithm, presents examples of solving problems and equations of various levels of complexity.

2.History of equations in integers

Diophantus - scientist - algebraist Ancient Greece, according to some sources, he lived until 364 AD. e. He specialized in solving problems in integers. Hence the name Diophantine equations. The most famous, solved by Diophantus, is the problem of "decomposing into two squares." Its equivalent is the well-known Pythagorean theorem. The life and work of Diophantus proceeded in Alexandria, he collected and solved known and invented new problems. Later he combined them in a large work called Arithmetic. Of the thirteen books that made up the Arithmetic, only six survived until the Middle Ages and became a source of inspiration for mathematicians of the Renaissance. Diophantus' Arithmetic is a collection of problems, each includes a solution and the necessary explanation. The collection includes a variety of problems, and their solution is often highly ingenious. Diophantus is only interested in positive integer and rational solutions. He calls irrational solutions "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. Many eminent mathematicians worked on a complete proof of the Great Theorem, and these efforts led to many results in modern number theory. It is believed that the theorem is in first place in terms of the number of incorrect proofs.

The remarkable French mathematician Pierre Fermat stated that the equation for an integer n ≥ 3 has no solutions in positive integers x, y, z (xyz = 0 is excluded by the positiveness of x, y, z. For the case n = 3, this theorem was tried in the X century proved by the Central Asian mathematician al-Khojandi, but his proof has not been preserved.Somewhat later, Fermat himself published a proof of a particular 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 prime n less than 100, with the possible exception of 37, 59, 67.

In the 1980s there was new approach to solving the problem. From the Mordell conjecture, proved by Faltings in 1983, it follows that the equation

for n > 3 can have only a finite number of coprime solutions.

The last but most important step in the proof of the theorem was taken in September 1994 by Wiles. His 130-page proof was published in 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 hypothesis (this assumption was proved by Ken Ribet with the participation of J.-P. Serra). Wiles published the first version of his proof in 1993 (after 7 years of hard work), but a serious gap was soon discovered in it; with the help of Richard Lawrence Taylor, the gap was quickly closed. The final version was published in 1995. March 15, 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 it is required to find only integer solutions of the equation. It can be seen 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 / f. 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 evenly divisible by 4, then

Because a=o and b=0, then x is any number

Because 7 is not even divisible by 10, then there are no solutions.

4. Way to enumerate options.

In the method of enumeration of options, it is necessary to take into account the signs of divisibility of numbers, to consider all possible options finite enumeration equality. This method can be used to solve these problems:

№1 Find the set of all pairs of natural numbers that are the solution of the equation 49x+69y=602

We express from the equation x =,

Because x and y are natural numbers, then x = ≥ 1, multiply the whole equation by 49 to get rid of the denominator:

Move 602 to the left side:

51y ≤ 553, express y, y= 10

A complete enumeration of options shows that the natural solutions of the equation are x=5, y=7.

Answer: (5,7).-

№2 Solve the problem

From the numbers 2, 4, 7, a three-digit number should be made, in which not a single number can be repeated more than two times.

Let's find the number of all three-digit numbers that start 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 the numbers 4 and 7: (442, 424, 422, 447, 474, 427, 472, 477).

(772, 774, 727, 747, 722, 744, 724, 742) - they are also 8 numbers each. There are only 24 numbers.

Answer: 24th.

5. Continued fraction and Euclid's algorithm

A continued fraction is an expression of an ordinary fraction in the form

where q 1 is an integer, and q 2 , … ,qn are natural numbers. Such an expression is called a continued (finite continued) fraction. There are finite and infinite continued fractions.

For rational numbers the continued fraction has end view. 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 of actions for this method of solving equations in integers.

Algorithm

1) Compile the ratio of the coefficients for unknowns in the form of a fraction

2) Convert expression to improper fraction

3) Select the integer part of an improper fraction

4) Replace a proper fraction with an equal fraction

5) Do 3.4 with the wrong fraction obtained in the denominator

6) Repeat 5 until the end 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 in the unknowns.

First of all, we select the integer part of the improper fraction; = 2 +

Replace a proper fraction with an equal fraction.

Where = 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 obtain

We got an expression called the final continued or 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 bring the resulting expression to a common denominator and discard it.

Whence 127∙9-52∙22+1=0. Comparing the obtained equality with the equation 127x- 52y+1 = 0, it follows that then x= 9, y= 22 is a 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....). , discard its last link and do calculations similar to those given above.

To prove this assumption, we will need some properties of continued fractions.

Consider an irreducible fraction. Denote by q 1 the quotient and by r 2 the remainder of dividing a by b. Then we get:

Then b=q 2 r 2 +r 3 ,

Similar

r 2 \u003d q 3 r 3 + r 4, ;

r 3 \u003d q 4 r 4 + r 5,;

………………………………..

Quantities q 1 , q 2 ,… are called incomplete quotients. The above process of forming incomplete quotients is called Euclid's algorithm. Remainders from division r 2 , r 3 ,…satisfy the inequalities

those. form a series of decreasing non-negative numbers.

Example #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 expansion of a fraction into a continued fraction

Having collapsed the penultimate fraction suitable for it into an ordinary

A particular solution of this equation has the form

X 0 \u003d (-1) 4300 ∙ 9 \u003d 2700, y 0 \u003d (-1) 5300 ∙ 8 \u003d -2400,

and the general is given by the formula

x=2700-19k, y=-2400+17k.

whence we obtain the condition on the parameter k

Those. k=142, x=2, y=14. .

6. Factoring method

The method of enumeration of options is an inconvenient way, 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 mathematics and in higher mathematics.

The essence consists in identical transformation. The meaning of any identical transformation is to write an expression in a different form while preserving its essence. Consider examples of the application of this method.

№1 Solve the equation in integers y 3 - x 3 = 91.

Using the abbreviated multiplication formulas, we decompose the right side of the equation into factors:

(y - x)(y 2 + xy + x 2) = 91

We write out all the divisors of the number 91: ± 1; ± 7; ± 13; ±91

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 the set of systems of equations:

Having solved the systems, we select those roots that are integers.

We get 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 that satisfy the equation x 2 -y 2 = 69

We factorize the left side of the equation and write the equation as

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, by solving which we can find the desired numbers:

Having expressed 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 \u003d xy in integers:

We write the equation in the form

Let's factorize the left side of the equation. 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.

We factorize the left side of the equation and divide both sides of the equation by 3, as a result we get the 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 10 in the product, will not equal 0. Therefore, the original equation has no solutions in integers.

7. Method of residuals

The main task of the method is to find the remainder of the division of both parts of the equation by an integer, based on the results obtained. Often the information obtained reduces the possibilities of the solution sets of the equation. Consider examples:

№1 Prove that the equation x 2 = 3y + 2 has no solutions in integers.

Proof.

Consider the case where x, y ∈ N. Consider the remainders of both sides 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 conclude that there is no solution to this equation in natural numbers.

Consider the case when one of the numbers is equal to 0. Then, obviously, there are no solutions in integers.

The case when y is a negative integer has no solutions, because the right side will be negative and the left side positive.

The case when x is a negative integer also has no solutions, because falls under one of the cases considered earlier due to the fact that (-x) 2 = (x) 2 .

It turns out that the indicated equation has no solutions in integers, which was required to be proved.

№2 Solve in integers 3 X = 1 + y 2 .

It is not difficult to see that (0; 0) is the solution of this equation. It remains to prove that the equation has no other integer roots.

Consider the cases:

1) If x∈N, y∈N, then Z 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 positive integers

values ​​of 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 the equation 2x 2 -2xy+9x+y=2 in integers:

Let us express from the equation the unknown that enters it only to the first degree, that is, the variable y:

2x 2 + 9x-2 = 2xy-y, whence

We select the integer part of the fraction using the rule for dividing a polynomial by a polynomial "angle". We get:

Obviously, a 2x-1 difference can only take on the values ​​-3, -1, 1, and 3.

It remains to enumerate these four cases, as a result of which we obtain 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 ones with respect to one of the variables

№1 Solve the equation 5x in integers 2 +5y 2 + 8xy+2y-2x +2=0

This equation can be solved by the factorization method, however, this method, as applied to this equation, is quite laborious. Let's consider a more rational way.

We write the equation in the form of a quadratic 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 in terms of 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 (foreign roots)

Answer: (25;3)

In addition to tasks to solve the 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 coprime.

№3 Prove that the equation x 2 - 3y = 17 has no integer solutions.

Proof:

Let x; y - solutions of the equation

x 2 \u003d 3 (y + 6) -1 y ∈ Z then y+6 ∈ Z , so 3(y+6) is divisible by 3, hence 3(y+6)-1 is not divisible by 3, hence x 2 is not divisible by 3, hence x is not divisible by 3, so x = 3k±1, k ∈ Z.

Substitute this into the original equation.

We got a contradiction. This means that the equation has no entire solutions, which was required to be proved.

10.Peak Formula

Pick's formula was discovered by the Austrian mathematician Georg Pick in 1899. The formula is related to equations in integers in that only integer nodes are taken from polygons, as well as integers in equations.

Using this formula, you can find the area of ​​\u200b\u200ba figure built on a sheet in a cell (triangle, square, trapezoid, rectangle, polygon).

In this formula, we will find integer points inside the polygon and on its border.

In the tasks that will be on the exam, there is a whole group of tasks in which a polygon is given built on a sheet in a cell and there is a question about finding the area. The cell scale is one square centimeter.

Example #1

M - the number of nodes on the border of the triangle (on the sides and vertices)

N is the number of nodes inside the triangle.

*Under "knots" we mean the intersection of lines. Find the area of ​​the triangle:

Note the nodes:

M = 15 (indicated in red)

N = 34 (marked in blue)

Example #2

Find the area of ​​the polygon: Note the nodes:

M = 14 (indicated in red)

N = 43 (marked 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.

We will 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 in terms of another unknown:

2) Select the integer part: Obviously, x will be integer if the expression turns out to be integer, which, in turn, will take place 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 already with respect to the variable y, arguing exactly the same as in paragraphs 1, 2: Selecting the integer part, we get:

5) Arguing 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 over (we continue the process until there are no fractions left in the expression for the next variable).

7) Now you need to "go up". 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 of the original equation in integers.

Thus, the descent method involves first sequential expression of one variable through another, until there are no fractions left in the representation of the variable, and then, sequential “ascent” 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 of solving them were known to me. In the course of research, I managed to find and describe little-known ways to solve 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 / Grade 10, Grade 11 / / M., “Prosveshchenie”, 1998;

2. A. F. Ivanov et al., Mathematics. Educational and training materials for preparing for the exam // Voronezh, GOUVPO VSTU, 2007

3. A. O. Gel’fond, Mathematics, number theory// Solving equations in integers// LIBROCOM Book House

Internet resources:

4. Demo Options control measuring materials of the unified state exam in mathematics http://fipi.ru/

5. Examples of solutions of equations in integers http://reshuege.ru

6. Examples of solutions of 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 Equationshttp://dok.opredelim.com/docs/index-1732.html

10. History of Diophantus http://www.studfiles.ru/preview/4518769/

1.3 Ways to solve equations

When solving equations in integer and natural numbers, we can conditionally distinguish following methods:

1. A way to enumerate options.

2. Euclid's algorithm.

3. Continued fractions.

4. Method of factorization.

5. Solving equations in integers as square ones with respect to some variable.

6. Method of residuals.

7. Method of infinite descent.

Chapter 2

1. Examples of solving equations.

2.1 Euclid's algorithm.

Task 1 . Solve the equation in integers 407 X – 2816y = 33.

Let's use the compiled algorithm.

1. Using the Euclid 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 to get Equation 37 X – 256y= 3, and (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, hence the pair of numbers x 0= – 83 and at 0= – 12 is the solution of Equation 37 X – 256y = 3.

4. Write down the general formula for solutions to the original equation

where t- any integer.

2.2 Way to enumerate options.

Task 2. Rabbits and pheasants sit in a cage, they have 18 legs in total. Find out how many of those and others 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.

Express at through X : y \u003d 9 - 2x.

Thus, the problem has four solutions.

Answer: (1; 7), (2; 5), (3; 3), (4; 1).

2.3 Factoring method.

The enumeration of options when finding natural solutions to an equation with two variables turns out to be very laborious. Also, if the equation is whole solutions, it is impossible to enumerate them, since there are an infinite number of such solutions. Therefore, we will show one more trick - factorization method.

Task 3. Solve the equation in integersy 3 - x 3 = 91.

Solution. 1) Using the abbreviated multiplication formulas, we decompose the right side of the equation into factors:

(y - x)(y 2 + xy + x 2) = 91……………………….(1)

2) Write out all the divisors of the number 91: ± 1; ± 7; ± 13; ±91

3) We conduct research. Note that for any integer 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 get: the first system has solutions (5; 6), (-6; -5); third (-3; 4),(-4; 3); the second and fourth solutions in integers do not have.

Answer: equation (1) has four solutions (5; 6); (-6; -5); (-3; 4); (-4;3).

Task 4. Find all pairs of natural numbers that satisfy the equation

Solution. We factorize the left side of the equation and write the equation as

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. Given that

The first system has a solution

Answer:

Task 5. Solve the equation in integers:

Solution. We write the equation in the form

Let's factorize the left side of the equation. 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 the solution x=2, y=2, and the second system has the solution x=0, y=0.

Answer:

Task 6. Solve the equation in integers

Solution. We write this equation in the form

We decompose the left side of the equation into factors by the grouping method, we get

The product of two integers can equal 7 in the following cases:

7=1 7=7 1=-1 (-7)=-7 (-1). Thus, we get four systems:

The solution of 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

The last video was devoted to linear equations containing two variables. We have considered the main properties of such expressions, the possibilities of their transformation and solution, as well as graphic display dependencies between two variables.

It is known that the vast majority of these equations have a set of answers, always represented by a pair of numbers. This pair is x and y values. Consider the possible variants of the roots of the equation of the following form:

Obviously, the pair (4, 6) can be the roots of this equation:

Or fractions 1/5 and 1/3:

5(1/5) - 3(1/3) = 2

In both cases, the correct equality is obtained, which means that both pairs of roots are acceptable as a solution to the presented equation. But at the same time, one pair is fractions, and the second is represented by integers. The roots of equations with two variables that have values ​​in integers are called integers.
Quite often in mathematics there are problems that require integer solutions of such equations. On the other hand, some variations like:

do not have a whole numerical solutions generally. Since for any integer values ​​of x and y, you get an integer general expression the left side (2x + 3y), which cannot be equal to a fraction in any way - that is, the principle of equality conservation will be violated.
Consider possible solutions to the equation:

Let's translate it into a dependency form using the transfer through the equal sign and identical transformations:

It is quite obvious that the equality of the form is preserved:

Where n is any natural number, which may well be an integer in value. That is, the equation 7x - y \u003d -1 has a set of integer solutions. Let's check any integers as x:

x = -3; y = -26

We already know the general abstract formula for defining any linear equation with two variables:

Where x and y are variables, a and b are coefficients of variables, and c is a free term. Any equation similar to linear expressions with x and y can be reduced to such an abstract form by equivalent transformations. Detailed study general formula makes it easy to identify some patterns in terms of the presence of integer solutions. So, if some equation of the form is given:

In which the free term is a fraction, then the roots of the equation cannot be integral numerical expressions. The sum or difference of two integers, according to the law of elementary algebra, cannot be equal to a fractional expression.

Due to the large number possible solutions, the roots of equations with two variables sometimes take the form not of a pair of individual numbers, but of a pair of two individual formulas - for x and for y. For example, let's solve the equation:

To do this, we need to make a number of transformations. Let's break the monomial 20x into the identical sum 18x + 2x:

20x = 18x + 2x

18x + 2x + 3y = 10

We group monomials that have multiple numerical coefficients. It is worth noting that the variable x must be divided into a sum so that x would be obtained with a coefficient as large as possible and a multiple of the numerical coefficient of the variable y. Since in our example at y there is a triple, then we break x with the maximum allowable coefficient, a multiple of three. After grouping, we take out the common multiple factor:

18x + 2x + 3y = 10

18x + 3y + 2x = 10

3(6x + y) + 2x = 10

Let the expression in brackets (6x + y) be equal to some variable c, then:

3(6x + y) + 2x = 10

We split the value of the variable c according to the same principle as we split the coefficient for x. In this case, we need to choose a certain number that will be a multiple of two (the value at 2x), but not more than three. Obviously it will be like this:

2s + s + 2x = 10

We make the same changes:

2s + s + 2x = 10

2(c + x) + c = 10

Let's denote the contents of the brackets as n, then:

2(c + x) + c = 10

We substitute the resulting equality instead of with:

3(10 - 2n) + 2x = 10

And we solve the resulting equation for the variable x:

3(10 - 2n) + 2x = 10

30 - 6n + 2x = 10

2x \u003d 10 + 6n - 30

It is appropriate to write:

6x + y \u003d n - x

We substitute the formula we know for x to calculate y:

6x + y \u003d n - x

6(- 10 + 3n) + y = n - (- 10 + 3n)

60 + 18n + y = n + 10 - 3n

y \u003d n + 10 - 3n + 60 - 18n

The roots of the equation 20x + 3y = 10 are two expressions of the form:

Where n is any integer - 0, 1, 2, etc. Thus, to describe the whole variety of possible integer solutions, the easiest way is to calculate some formulas for quickly calculating x and y. Substituting any expressions n in these formulas, you can easily get the desired pair of numbers.