Equation x2 y2. Solving equations with two variables
1. Systems of linear equations with a parameter
Systems of linear equations with a parameter are solved by the same basic methods as conventional systems of equations: the substitution method, the method of adding equations, and the graphical method. Knowing the graphical interpretation of linear systems makes it easy to answer the question about the number of roots and their existence.
Example 1
Find all values for the parameter a for which the system of equations has no solutions.
(x + (a 2 - 3) y \u003d a,
(x + y = 2.
Solution.
Let's look at several ways to solve this problem.
1 way. We use the property: the system has no solutions if the ratio of the coefficients in front of x is equal to the ratio of the coefficients in front of y, but not equal to the ratio of free terms (a/a 1 = b/b 1 ≠ c/c 1). Then we have:
1/1 \u003d (a 2 - 3) / 1 ≠ a / 2 or a system
(and 2 - 3 = 1,
(a ≠ 2.
From the first equation a 2 \u003d 4, therefore, taking into account the condition that a ≠ 2, we get the answer.
Answer: a = -2.
2 way. We solve by the substitution method.
(2 - y + (a 2 - 3) y \u003d a,
(x = 2 - y,
((a 2 - 3) y - y \u003d a - 2,
(x = 2 - y.
After taking the common factor y out of brackets in the first equation, we get:
((a 2 - 4) y \u003d a - 2,
(x = 2 - y.
The system has no solutions if the first equation has no solutions, that is
(and 2 - 4 = 0,
(a - 2 ≠ 0.
It is obvious that a = ±2, but taking into account the second condition, only the answer with a minus is given.
Answer: a = -2.
Example 2
Find all values for the parameter a for which the system of equations has an infinite number of solutions.
(8x + ay = 2,
(ax + 2y = 1.
Solution.
By property, if the ratio of the coefficients at x and y is the same, and is equal to the ratio of the free members of the system, then it has an infinite number of solutions (i.e., a / a 1 \u003d b / b 1 \u003d c / c 1). Hence 8/a = a/2 = 2/1. Solving each of the equations obtained, we find that a \u003d 4 is the answer in this example.
Answer: a = 4.
2. Systems of rational equations with a parameter
Example 3
(3|x| + y = 2,
(|x| + 2y = a.
Solution.
Multiply the first equation of the system by 2:
(6|x| + 2y = 4,
(|x| + 2y = a.
Subtract the second equation from the first, we get 5|х| = 4 – a. This equation will have a unique solution for a = 4. In other cases, this equation will have two solutions (for a< 4) или ни одного (при а > 4).
Answer: a = 4.
Example 4
Find all values of the parameter a for which the system of equations has a unique solution.
(x + y = a,
(y - x 2 \u003d 1.
Solution.
We will solve this system using the graphical method. So, the graph of the second equation of the system is a parabola, lifted up along the Oy axis by one unit segment. The first equation defines the set of lines parallel to the line y = -x (picture 1). The figure clearly shows that the system has a solution if the straight line y \u003d -x + a is tangent to the parabola at the point with coordinates (-0.5; 1.25). Substituting these coordinates into the equation of a straight line instead of x and y, we find the value of the parameter a: 
1.25 = 0.5 + a;
Answer: a = 0.75.
Example 5
Using the substitution method, find out at what value of the parameter a, the system has a unique solution.
(ax - y \u003d a + 1,
(ax + (a + 2)y = 2.
Solution.
Express y from the first equation and substitute it into the second:
(y \u003d ah - a - 1,
(ax + (a + 2) (ax - a - 1) = 2.
We bring the second equation to the form kx = b, which will have a unique solution for k ≠ 0. We have:
ax + a 2 x - a 2 - a + 2ax - 2a - 2 \u003d 2;
a 2 x + 3ax \u003d 2 + a 2 + 3a + 2.
The square trinomial a 2 + 3a + 2 can be represented as a product of brackets
(a + 2)(a + 1), and on the left we take x out of brackets:
(a 2 + 3a) x \u003d 2 + (a + 2) (a + 1).
Obviously, a 2 + 3a must not be equal to zero, therefore,
a 2 + 3a ≠ 0, a(a + 3) ≠ 0, which means a ≠ 0 and ≠ -3.
Answer: a ≠ 0; ≠ -3.
Example 6
Using the graphical solution method, determine at what value of the parameter a, the system has a unique solution.
(x 2 + y 2 = 9,
(y - |x| = a.
Solution.
Based on the condition, we build a circle with a center at the origin of coordinates and a radius of 3 unit segments, it is this circle that sets the first equation of the system
x 2 + y 2 = 9. The second equation of the system (y = |x| + a) is a broken line. By using figure 2 we consider all possible cases of its location relative to the circle. It is easy to see that a = 3. 
Answer: a = 3.
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Instruction
Substitution method Express one variable and substitute it into another equation. You can express any variable you like. For example, express "y" from the second equation:
x-y=2 => y=x-2 Then plug everything into the first equation:
2x+(x-2)=10 Move everything without x to the right side and count:
2x+x=10+2
3x=12 Next, for "x, divide both sides of the equation by 3:
x=4. So, you have found "x. Find "at. To do this, substitute "x" into the equation from which you expressed "y:
y=x-2=4-2=2
y=2.
Make a check. To do this, substitute the resulting values into the equations:
2*4+2=10
4-2=2
Unknown found correctly!
How to add or subtract equations Get rid of any variable at once. In our case, this is easier to do with "y.
Since in the equation “y has a sign” + , and in the second “-”, then you can perform an addition operation, i.e. We add the left side to the left, and the right side to the right:
2x+y+(x-y)=10+2Convert:
2x+y+x-y=10+2
3x=12
x=4 Substitute "x" into any equation and find "y:
2*4+y=10
8+y=10
y=10-8
y=2 By the 1st method, you can check that the roots are found correctly.
If there are no clearly defined variables, then it is necessary to slightly transform the equations.
In the first equation we have "2x", and in the second just "x. In order for x to be reduced when adding or subtracting, multiply the second equation by 2:
x-y=2
2x-2y=4 Then subtract the second equation from the first equation:
2x+y-(2x-2y)=10-4
2x+y-2x+2y=6
3y=6
find y \u003d 2 "x by expressing from any equation, i.e.
x=4
Related videos
When solving differential equations, the argument x (or time t in physical problems) is not always explicitly available. Nevertheless, this is a simplified special case of setting a differential equation, which often helps to simplify the search for its integral.
Instruction
Consider a physics problem that leads to a differential equation that lacks an argument t. This is the problem of vibrations of mass m, suspended on a thread of length r, located in a vertical plane. The equation of motion of the pendulum is required if the initial one was stationary and deviated from the equilibrium state by an angle α. Forces should be neglected (see Fig. 1a).
Solution. A mathematical pendulum is a material point suspended on a weightless and inextensible thread at point O. Two forces act on the point: gravity G \u003d mg and thread tension N. Both of these forces lie in the vertical plane. Therefore, to solve the problem, it is possible to apply the equation of rotational motion of a point around a horizontal axis passing through point O. The equation of rotational motion of a body has the form shown in fig. 1b. In this case, I is the moment of inertia of the material point; j is the angle of rotation of the thread together with the point, counted from the vertical axis counterclockwise; M is the moment of forces applied to the material point.
Calculate these quantities. I=mr^2, M=M(G)+M(N). But M(N)=0, since the line of action of the force passes through the point O. M(G)=-mgrsinj. The "-" sign means that the moment of force is directed in the direction opposite to the movement. Substitute the moment of inertia and the moment of force into the equation of motion and get the equation shown in Fig. 1s. By reducing the mass, a relation arises (see Fig. 1d). There is no t argument here.
Solving equations in integers is one of the oldest mathematical problems. Already at the beginning of the 2nd millennium BC. e. The Babylonians knew how to solve systems of such equations with two variables. This area of mathematics reached its greatest prosperity in ancient Greece. The main source for us is the "Arithmetic" of Diophantus, which contains various types of equations. In it, Diophantus (after his name and the name of the equations - Diophantine equations) anticipates a number of methods for studying equations of the 2nd and 3rd degrees, which developed only in the 19th century.
The simplest Diophantine equations ax + y = 1 (equation with two variables, first degree) x2 + y2 = z2 (equation with three variables, second degree)
Algebraic equations have been studied most fully; their solution was one of the most important problems in algebra in the 16th and 17th centuries.
By the beginning of the 19th century, the works of P. Fermat, L. Euler, K. Gauss investigated a Diophantine equation of the form: ax2 + bxy + cy2 + dx + ey + f = 0, where a, b, c, d, e, f are numbers; x, y are unknown variables.
This is a 2nd degree equation with two unknowns.
K. Gauss built a general theory of quadratic forms, which is the basis for solving certain types of equations with two variables (Diophantine equations). There are a large number of specific Diophantine equations that can be solved by elementary methods. /p>
theoretical material.
In this part of the work, the basic mathematical concepts will be described, the definitions of terms will be given, the decomposition theorem will be formulated using the method of indefinite coefficients, which were studied and considered when solving equations with two variables.
Definition 1: An equation of the form ax2 + bxy + cy2 + dx + ey + f = 0, where a, b, c, d, e, f are numbers; x, y unknown variables is called a second-degree equation with two variables.
In the school course of mathematics, the quadratic equation ax2 + inx + c \u003d 0 is studied, where a, b, c of the number x is a variable, with one variable. There are many ways to solve such an equation:
1. Finding roots using the discriminant;
2. Finding roots for an even coefficient in (according to D1 =);
3. Finding roots by Vieta's theorem;
4. Finding the roots using the selection of the full square of the binomial.
Solving an equation means finding all its roots or proving that there are none.
Definition 2: The root of an equation is a number that, when substituted into the equation, forms a true equality.
Definition 3: The solution of an equation with two variables is called a pair of numbers (x, y), when substituting them into the equation, it turns into a true equality.
The process of searching for solutions to an equation very often usually consists in replacing the equation with an equivalent equation, but simpler in solution. Such equations are called equivalent.
Definition 4: Two equations are said to be equivalent if each solution to one equation is a solution to the other equation, and vice versa, and both equations are considered in the same area.
To solve equations with two variables, the theorem on the expansion of the equation into a sum of perfect squares is used (by the method of indefinite coefficients).
For the second order equation ax2 + bxy + cy2 + dx + ey + f = 0 (1) there is a decomposition a(x + py + q)2 + r(y + s)2 + h (2)
Let us formulate the conditions under which expansion (2) takes place for equation (1) of two variables.
Theorem: If the coefficients a, c, c of equation (1) satisfy the conditions a0 and 4av - c20, then expansion (2) is determined in a unique way.
In other words, equation (1) with two variables can be reduced to the form (2) using the method of indefinite coefficients, if the conditions of the theorem are satisfied.
Let's look at an example of how the method of indefinite coefficients is implemented.
METHOD #1. Solve the equation by the method of indeterminate coefficients
2 x2 + y2 + 2xy + 2x + 1 = 0.
1. Let's check the fulfillment of the conditions of the theorem, a=2, b=1, c=2, so a=2,4av - c2= 4∙2∙1- 22= 40.
2. The conditions of the theorem are satisfied, and can be expanded by formula (2).
3. 2 x2 + y2 + 2xy + 2x + 1 = 2(x + py + q)2 + r(y + s)2 + h, based on the conditions of the theorem, both parts of the identity are equivalent. Simplify the right side of the identity.
4. 2(x + py + q)2 + r(y +s)2 +h =
2(x2 + p2y2 + q2 + 2pxy + 2pqy + 2qx) + r(y2 + 2sy + s2) + h =
2x2+ 2p2y2 + 2q2 + 4pxy + 4pqy + 4qx + ry2 + 2rsy + rs2 + h =
X2(2) + y2(2p2 + r) + xy(4p) + x(4q) + y(4pq + 2rs) + (2q2 + rs2 + h).
5. Equate the coefficients for the same variables with their powers.
x2 2 = 2 y21 = 2p2 + r) xy2 = 4p x2 = 4q y0 = 4pq + 2rs x01 = 2q2 + rs2 + h
6. Get a system of equations, solve it and find the values of the coefficients.
7. Substitute the coefficients in (2), then the equation will take the form
2 x2 + y2 + 2xy + 2x + 1 \u003d 2 (x + 0.5y + 0.5) 2 + 0.5 (y -1) 2 + 0
Thus, the original equation is equivalent to the equation
2(x + 0.5y + 0.5)2 + 0.5(y -1)2 = 0 (3), this equation is equivalent to a system of two linear equations.
Answer: (-1; 1).
If you pay attention to the type of decomposition (3), you can see that it is identical in form to the extraction of a full square from a quadratic equation with one variable: ax2 + inx + c = a(x +)2 +.
Let's apply this trick to solving an equation with two variables. Let's solve with the help of selection of the full square the quadratic equation with two variables already solved using the theorem.
METHOD #2: Solve the equation 2x2 + y2 + 2xy + 2x +1= 0.
Solution: 1. We represent 2x2 as the sum of two terms x2 + x2 + y2 + 2xy + 2x + 1 = 0.
2. We group the terms in such a way that we can collapse according to the full square formula.
(x2 + y2 + 2xy) + (x2 + 2x + 1) = 0.
3. Select the full squares from the expressions in brackets.
(x + y)2 + (x + 1)2 = 0.
4. This equation is equivalent to a system of linear equations.
Answer: (-1;1).
If we compare the results, we can see that the equation solved by method No. 1 using the theorem and the method of indefinite coefficients and the equation solved by method No. 2 using the selection of a full square have the same roots.
Conclusion: A quadratic equation with two variables can be expanded into a sum of squares in two ways:
➢ The first method is the method of indeterminate coefficients, which is based on the theorem and expansion (2).
➢ The second way is with the help of identical transformations, which make it possible to select consecutively complete squares.
Of course, when solving problems, the second method is preferable, since it does not require memorizing expansion (2) and conditions.
This method can also be applied to quadratic equations with three variables. The selection of the full square in such equations is more laborious. I will be doing this kind of transformation next year.
It is interesting to note that a function that looks like: f(x, y)= ax2 + bxy + cy2 + dx + ey + f is called a quadratic function of two variables. Quadratic functions play an important role in various branches of mathematics:
In mathematical programming (quadratic programming)
In linear algebra and geometry (quadratic forms)
In the theory of differential equations (reducing a second-order linear equation to a canonical form).
When solving these various problems, one has, in fact, to apply the procedure for extracting the full square from a quadratic equation (one, two or more variables).
Lines whose equations are described by a quadratic equation of two variables are called curves of the second order.
This circle, ellipse, hyperbola.
When plotting these curves, the method of successive selection of the full square is also used.
Let's consider how the method of successive selection of a full square works on specific examples.
Practical part.
Solve equations using the method of successive selection of the full square.
1. 2x2 + y2 + 2xy + 2x + 1 = 0; x2 + x2 + y2 + 2xy + 2x + 1 = 0;
(x + 1)2 + (x + y)2 = 0;
Answer: (-1; 1).
2. x2 + 5y2 + 2xy + 4y + 1 = 0; x2 + 4y2 + y2 + 2xy + 4y + 1 = 0;
(x + y)2 + (2y + 1)2 = 0;
Answer: (0.5; - 0.5).
3. 3x2 + 4y2 - 6xy - 2y + 1 = 0;
3x2 + 3y2 + y2 - 6xy - 2y +1 = 0;
3x2 + 3y2 - 6xy + y2 -2y +1 = 0;
3(x2 - 2xy + y2) + y2 - 2y + 1 = 0;
3(x2 - 2xy + y2)+(y2 - 2y + 1)=0;
3(x-y)2 + (y-1)2 = 0;
Answer: (-1; 1).
Solve Equations:
1. 2x2 + 3y2 - 4xy + 6y +9 = 0
(bring to the form: 2(x-y)2 + (y +3)2 = 0)
Answer: (-3; -3)
2. - 3x2 - 2y2 - 6xy -2y + 1=0
(bring to the form: -3 (x + y) 2 + (y -1) 2 \u003d 0)
Answer: (-1; 1)
3. x2 + 3y2 + 2xy + 28y +98 = 0
(bring to the form: (x + y) 2 + 2 (y + 7) 2 \u003d 0)
Answer: (7; -7)
Conclusion.
In this scientific work, equations with two variables of the second degree were studied, methods for solving them were considered. The task is completed, a shorter solution method is formulated and described, based on selecting a full square and replacing the equation with an equivalent system of equations, as a result, the procedure for finding the roots of an equation with two variables is simplified.
An important point of the work is that the method under consideration is used in solving various mathematical problems associated with a quadratic function, constructing second-order curves, and finding the largest (smallest) value of expressions.
Thus, the technique of expanding a second-order equation with two variables into a sum of squares has the most numerous applications in mathematics.
Indefinite equations in natural numbers.
State Educational Institution "Rechitsa District Lyceum"
Prepared by: .
Supervisor: .
Introduction
1.Solving equations by factoring method…………4
2. Solving equations with two variables (discriminant method)……………………………………………………………………….11
3. Residue method ............................................... ...................................13
4. Method of "infinite descent" .............................................. ..............fifteen
5.Sampling method…………………………………………………………...16
Conclusion................................................. ........................................eighteen
Introduction
I am Slava, I study at the Rechitsa District Lyceum, a student of the 10th grade.
Everything starts with an idea! I was asked to solve an equation with three unknowns 29x + 30y + 31 z =366. Now I regard this equation as a task - a joke, but for the first time I broke my head. For me, this equation has become kind of undefined, how to solve it, in what way.
Under indefinite equations we must understand that these are equations containing more than one unknown. Usually, people who solve these equations look for solutions in integers.
Solving indefinite equations is a very exciting and informative activity that contributes to the formation of ingenuity, observation, attentiveness in students, as well as the development of memory and orientation, the ability to think logically, analyze, compare and generalize. I have not yet found a general technique, but now I will tell you about some methods for solving such equations in natural numbers.
This topic is not fully covered in existing mathematics textbooks, and problems are offered at olympiads and at centralized testing. This interested and fascinated me so much that when solving various equations and problems, I gathered a whole collection of my own solutions, which we divided with the teacher according to the methods and methods of solving. So what is my purpose of work?
My goal analyze solutions of equations with several variables on the set of natural numbers.
To begin with, we will consider practical problems, and then we will move on to solving equations.
What is the length of the sides of a rectangle if its perimeter is numerically equal to its area?
P=2(x+y),
S = xy, x€ N and y€ N
P=S
2x+2y=xy font-size:14.0pt;line-height: 150%;font-family:" times new roman>+font-size:14.0pt;line-height: 150%;font-family:" times new roman>=font-size:14.0pt;line-height:150%;font-family:" times new roman position:relative>font-size:14.0pt;line-height: 150%;font-family:" times new roman> +font-size:14.0pt;line-height: 150%;font-family:" times new roman> =font-size:14.0pt;line-height:150%;font-family:" times new roman>Answer: (4:4); (3:6); (6:3).
Find ways to pay 47 rubles, if only three and five-ruble bills can be used for this.
Solution
5x+3y=47
x=1, y=14
x=1 – 3K, y= 14+5K, K€ Z
Natural values of x and y correspond to K= 0, -1, -2;
(1:14) (4:9) (7:4)
Joke task
Prove that there is a solution to the equation 29x+30y+31 z=336 in natural numbers.
Proof
A leap year has 366 days and one month has 29 days, four months have 30 days,
7 months - 31 days.
The solution is three (1:4:7). This means that there is a solution to the equation in natural numbers.
1. Solving equations by factoring
1) Solve the equation x2-y2=91 in natural numbers
Solution
(x-y)(x+y)=91
Solution 8 systems
font-size:14.0pt; line-height:150%;font-family:" times new roman>x-y=1
x+y=91
(46:45)
font-size:14.0pt; line-height:150%;font-family:" times new roman>x-y=91
x+y=1
(46: -45)
x-y=13
x+y=7
(10: -3)
x-y = 7
x+y=13
(10:3)
x-y = -1
x+y= -91
(-46: 45)
x-y = -91
x+y= -1
(-46: -45)
x-y = -13
x+y= -7
(-10:3)
x-y font-size:14.0pt; line-height:150%;font-family:" times new roman>= -7
x+y= -13
(-10: -3)
Answer: ( 46:45):(10:3).
2) Solve the equation x3 + 91 \u003d y3 in natural numbers
Solution
(y-x)(y2+xy+x2)=91
91=1*91=91*1=13*7=7*13= (-1)*(-91)=(-7)*(-13)
Solution 8 systems
y-x=1
y2+xy+x2=91
(5:6)(-6: -5)
font-size:14.0pt; line-height:150%;font-family:" times new roman>y-x= 91
y2+xy+x2= 1
y-x=13
y2+xy+x2=7
has no solutions in integers
y-x=7
y2+xy+x2=91
(-3: 4)(-4: 3)
The remaining 4 systems do not have solutions in integers. The condition is satisfied by one solution.
Answer: (5:6).
3) Solve the equation xy=x+y in natural numbers
Solution
xy-x-y+1=1
x(y-1)-(y-1)=1
(y-1)(x-1)=1
1= 1*1=(-1)*(-1)
Solution 2 systems
font-size:14.0pt; line-height:150%;font-family:" times new roman>y-1= -1
x-1= -1
(0:0)
font-size:14.0pt; line-height:150%;font-family:" times new roman>y-1=1
x-1=1
(2:2)
Answer: (2:2).
4) Solve the equation 2x2+5xy-12y2=28 in natural numbers
Solution
2x2-3xy+8xy-12y2=28
(2x-3y)(x+4y)=28
x;y - natural numbers; (x+4y)€ N
(x+4y)≥5
font-size:14.0pt; line-height:150%;font-family:" times new roman>2x-3y=1
x+4y=28
(8:5)
font-size:14.0pt; line-height:150%;font-family:"times new roman>2x-3y=4
x + 4y = 7
2x-3y=2
x+4y=14
no solutions in natural numbers
Answer: (8:5).
5) solve the equation 2xy=x2+2y in natural numbers
Solution
x2-2xy+2y=0
(x2-2xy+y2)-y2+2y-1+1=0
(x-y)2-(y-1)2= -1
(x-y-y+1)(x-y+y-1)= -1
(x-2y+1)(x-1)= -1
x-2y+1=-1
x-1= 1
(2:2)
x-2y+1=1
x-1= -1
no solutions in natural numbers
Answer: (2:2).
6) solve the equation xatz-3 xy-2 xz+ yz+6 x-3 y-2 z= -4 in natural numbers
Solution
xy(z -3)-2 x (z -3)+ y (z -3)-2 z +4=0
xy(z -3)-2 x (z -3)+ y (z -3)-2 z +6-2=0
xy(z -3)-2 x (z -3)+ y (z -3)-2(z -3)=2
(z-3)(xy-2x+y-2)=2
(z-3)(x(y-2)+(y-2))=2
(z-3)(x+1)(y-2)=2
Solution 6 systems
z -3= 1
x+1=1
y-2=2
(0 : 4 : 4 )
z-3= -1
x+1=-1
y-2= 2
(- 2: 4 : 2 )
EN-US" style="font-size: 14.0pt;line-height:150%;font-family:" times new roman>z-3= 1
x+1=2
y-2=1
(1 : 3 : 4 )
z-3=2
x+1=1
y-2=1
(0 :3: 5 )
z-3= -1
x +1 = 2
y-2=-1
(1:1:2)
z-3=2
x +1= -1
y -2= -1
(-2:1:5)
Answer: (1:3:4).
Consider a more complex equation for me.
7) Solve the equation x2-4xy-5y2=1996 in natural numbers
Solution
(x2-4xy+4y2)-9y2=1996
(x-2y)2-9y2=1996
(x-5y)(x+5y)=1996
1996=1*1996= -1*(-1996)=2*998= (-2)*(-998)=4*499= -4*(-499)
x € N , y € N ; (x+y) € N ; (x+y)>1
x-5y=1
x+y=1996
no solutions
font-size:14.0pt; line-height:150%;font-family:" times new roman>x-5y=499
x+y=4
no solutions
font-size:14.0pt; line-height:150%;font-family:" times new roman>x-5y=4
x+y=499
no solutions
x-5y=2
x+y=998
(832:166)
x-5y=988
x+y=2
no solutions
Answer: x=832, y=166.
Let's conclude:when solving equations by factoring, abbreviated multiplication formulas, grouping method, full square selection method are used .
2. Solving equations with two variables (discriminant method)
1) Solve the equation 5x2 + 5y2 + 8xy + 2y-2x + 2 \u003d 0 in natural numbers
Solution
5x2+(8y-2)x+5y2+2y+2=0
D \u003d (8y - 2) 2 - 4 * 5 * (5y2 + 2y + 2) \u003d 4 ((4y - 1) 2 -5 * (5y2 + 2y + 2))
x1.2= font-size:14.0pt;line-height: 150%;font-family:" times new roman>=font-size:14.0pt;line-height: 150%;font-family:" times new roman>
D=0, font-size:14.0pt;line-height: 150%;font-family:"times new roman>=0
y=-1, x=1
Answer: there are no solutions.
2) Solve the equation 3(x2+xy+y2)=x+8y in natural numbers
Solution
3(x2+xy+y2)=x+8y
3x2+3(y-1)x+3y2-8y=0
D \u003d (3y-1) 2-4 * 3 (3y2-8y) \u003d 9y2-6y + 1-36y2 + 96y \u003d -27y2 + 90y + 1
D≥0, -27y2+90y+1≥0
font-size:14.0pt;line-height: 150%;font-family:" times new roman>≤y≤font-size:14.0pt;line-height:150%;font-family:" times new roman>y€ N , y=1, 2, 3. Going through these values, we have (1:1).
Answer: (1:1).
3) Solve the equation x4-y4-20x2+28y2=107 in natural numbers
Solution
We introduce a replacement: x2=a, y2=a;
a2-a2-20a+28a=107
a2-20a+28a-a2=0
a1.2=-10± +96 font-size:14.0pt;line-height:150%;font-family:" times new roman color:black>a2-20a+28a-a2-96=11
a1,2=10± font-size:14.0pt;line-height: 150%;font-family:" times new roman>= 10±font-size:14.0pt;line-height: 150%;font-family:" times new roman>= 10±(a-14)
a1=a-4, a2=24-a
The equation looks like:
(a-a+4)(a+a-24)=1
font-size:14.0pt; line-height:150%;font-family:" times new roman>x2-y2+4=1
x2+y2 – 24=11
there are no solutions in natural numbers;
x2 - y2+4=11
x2+y2 – 24=1
(4:3),(-4:-3),(-4:3), (4: -3)
font-size:14.0pt; line-height:150%;font-family:" times new roman>x2 - y2+4= -1
x2 + y2 - 24 = -11
(2:3),(-2: -3),(-2:3),(2: -3)
x2 - y2+4= -11
х2+y2 – 24= -1 no solutions in natural and integer numbersAnswer: (4:3),(2:3).
3. Residual method
When solving equations by the residual method, the following tasks are very often used:
A) What remainders can give when divided by 3 and 4?
It's very simple, when divided by 3 or 4, exact squares can give two possible remainders: 0 or 1.
B) What remainders can give exact cubes when divided by 7 and 9?
When divided by 7, they can give remainders: 0, 1, 6; and when dividing by 9: 0, 1, 8.
1) Solve the equation x2+y2=4 z-1 in natural numbers
Solution
x2+y2+1=4z
Consider what remainders can give when divided by 4, the left and right sides of this equation. When divided by 4, exact squares can give only two different remainders 0 and 1. Then x2 + y2 + 1 when divided by 4 give remainders 1, 2, 3, and 4 z divided without remainder.
Therefore, this equation has no solutions.
2) Solve the equation 1!+2!+3!+ …+x!= y2 in natural numbers
Solution
a) X=1, 1!=1, then y2=1, y=±1 (1:1)
b) x=3, 1!+2!+3!= 1+2+6= 9, i.e. y2= 9, y=±3 (3:3)
c) x=2, 1!+2!= 1+2= 3, y2=3, i.e. y=±font-size:14.0pt;line-height:150%; font-family:"times new roman>d)x=4, 1!+2!+3!+4!= 1+2+6+24=33, x=4 (none), y2=33
e) x≥5, 5!+6!+…+x!, imagine 10 n , n € N
1!+2!+3! +5!+…+x!=33+10n
A number ending in 3 means that it cannot be the square of an integer. Therefore, x≥5 has no solutions in natural numbers.
Answer:(3:3) and (1:1).
3) Prove that there are no solutions in natural numbers
x2-y3=7
z 2 – 2у2=1
Proof
Assume the system is solvable z 2 \u003d 2y2 + 1, z2 - odd number
z=2m+1
y2+2m2+2m , y2 is an even number, y = 2 n , n € N
x2=8n3 +7, that is x2 is an odd number and X odd, x = 2 r +1, n € N
Substitute X and at into the first equation,
2(r 2 + r -2n 3 )=3
It is not possible, since the left side of the equation is divisible by two, and the right one is not divisible, which means that our assumption is not true, that is, the system has no solutions in natural numbers.
4. Infinite Descent Method
We solve according to the following scheme:
Suppose that the equation has a solution, we are building a certain infinite process, while, according to the very meaning of the problem, this process should end at an even step.
1)Prove that the equation 8x4+4y4+2 z4 = t4 has no solutions in natural numbers
Proof
Assume that the equation has a solution in integers, then it follows that
t4 is an even number, then t is also even
t=2t1 , t1 € Z
8x4 + 4y4 + 2 z 4 \u003d 16t14
4x4 + 2y4 + z 4 \u003d 8t14
z 4 \u003d 8t14 - 4x4 - 2y4
z 4 is even, then z =2 z 1 , z 1 € Z
Substitute
4x4 + 2y4 + 16 z 4 \u003d 8t14
y4 \u003d 4t14 - 2x4 - 8 z 1 4
x is even, i.e. x=2x, x1€ Z , then
16х14 – 2 t 1 4 – 4 z 1 4 +8 y 1 4 =0
8x14+4y14+2 z 1 4 = t 1 4
So x, y, z , t – even numbers, then x1, y1, z1,t1 - even. Then x, y, z, t and x1, y1, z 1, t 1 are divisible by 2, that is, font-size:14.0pt;line-height:150%;font-family:" times new roman position:relative>font-size:14.0pt;line-height: 150%;font-family:" times new roman>,font-size:14.0pt;line-height: 150%;font-family:"times new roman>,font-size:14.0pt;line-height: 150%;font-family:" times new roman> andfont-size:14.0pt;line-height: 150%;font-family:"times new roman>,font-size:14.0pt;line-height: 150%;font-family:"times new roman>,font-size:14.0pt;line-height: 150%;font-family:"times new roman>,font-size:14.0pt;line-height: 150%;font-family:"times new roman>.
So, it turned out that the number satisfies the equation; are multiples of 2, and no matter how many times we divide them by 2, we will always get numbers that are multiples of 2. The only number that satisfies this condition is zero. But zero does not belong to the set of natural numbers.
5. Sample method
1) Find solutions to the equation font-size:14.0pt;line-height: 150%;font-family:" times new roman>+font-size:14.0pt;line-height: 150%;font-family:" times new roman>=font-size:14.0pt;line-height:150%;font-family:" times new roman>Solution
font-size:14.0pt;line-height: 150%;font-family:" times new roman>=font-size:14.0pt;line-height:150%;font-family:" times new roman>p(x+y)=xy
xy=px+py
xy-px-ru=0
xy-px-ru+p2=p2
x(y-r)-p(y-r)=p2
(y-p)(x-p)=p2
p2= ±p= ±1= ±p2
Solution 6 systems
font-size:14.0pt; line-height:150%;font-family:" times new roman>y-r=r
x-p = p
y=2p, x=2p
y-r = - r
x-p = - p
y=0, x=0
y-r=1
x-p=1
y=1+p, x=1+p
y-r= -1
x-p = -1
y=p-1, x=p-1
font-size:14.0pt; line-height:150%;font-family:" times new roman>y-p=p2
x-p = p2
y=p2+p, x= p2+p
font-size:14.0pt; line-height:150%;font-family:" times new roman>y-p= -p2
x-p = - p2
y=p-p2, x=p-p2
Answer:(2p:2p), ( 1+p:1+p), (p-1:p-1), (p2+p:p2+p), (p-p2:p-p2).
Conclusion
Usually, solutions of indefinite equations are sought in integers. Equations in which only integer solutions are sought are called diophantine.
I analyzed the solutions of equations with more than one unknown, on the set of natural numbers. Such equations are so diverse that there is hardly any way, an algorithm for solving them. The solution of such equations requires ingenuity and contributes to the acquisition of independent work skills in mathematics.
I solved the examples with the simplest methods. The simplest technique for solving such equations is to express one variable in terms of the rest, and we get an expression that we will investigate in order to find these variables for which it is natural (integer).
At the same time, the concepts and facts related to divisibility, such as prime and composite numbers, signs of divisibility, relatively prime numbers, etc.
Especially often used:
1) If a product is divisible by a prime number p, then at least one of its factors is divisible by p.
2) If the product is divisible by some number With and one of the factors is coprime with the number With, then the second factor is divisible by With.
