Try to prove that the binary quadratic indefinite equation 25X X 6X 95 Y Y Y has no positive integer

We can think of this equation as a quadratic equation about x.

25x*x-6x-95-y^2+y=0

According to Veda's theorem.

x1*x2=-(95+y^2-y)/25

Because it is a positive integer solution, x1*x2 is a positive integer.

So -(95+y 2-y) 25 must be a positive integer.

and 95+y 2-y<0

From the discriminant expression of the root, it can be seen that 95+y 2-y<0 has no solution.

Therefore, there is no positive integer y, and there is no x, so there is no solution to the original form.

The upstairs seems to have been thought out too simply.

25x 2-95=5 (5x 2-19) When x is an odd number, the left formula still has a factor of 2.

But my approach is a bit complicated.

After the shifting transformation, it can be obtained that (5x-y)(5x+y-1)=x+95, if there is a positive integer solution, a=x+95, b=5x-y, c=5x+y-1, it is easy to know that a, b, and c are all positive integers.

Then a=bc c+b+1=10(a-95), i.e., c+b+1=10(bc-95), multiply 10 on both sides to get (10b-1) (10c-1)=9511, since 9511 is a prime number, so one of 10b-1 and 10c-1 is 1

It is concluded that there is no integer solution to the contradiction.

Isn't the element decomposition method a number theory method?

The power of odd numbers is odd and the power of even numbers is even, and from the original equation: x and y have the same parity. ··

1. When x and y are even, let x 2a and y 2b, where a and b are integers.

Then the original equation becomes: 4a 2 32b 5 4, a 2 8b 5 1, a must be an odd number.

Let a 2c 1, where c is an integer.

Then the original equation becomes: (2c 1) 2 8b 5 1, 4c 2 4c 1 8b 5 1, 2c 2 2c 4b 5 1The left side of this equation is an even number and the right side is an odd number, which is obviously wrong.

x and y cannot be even at the same time. ··

2. When x and y are both odd numbers, let x 2a 1 and y 2b 1, where a and b are integers.

Then the original equation becomes: (2a 1) 2 (2b 1) 3(2b 1) 2 1,4a 2 4a 1 (2b 1) 3(4b 2 4b 1) 1,4a 2 4a 4(b 2 b)(2b 1) 3

The left side of this equation is an even number and the right side is an odd number, which is obviously wrong.

x and y cannot be odd at the same time. ··

From , , , x, y cannot be integers, and the original equation has no integer solution. ,4,Prove: The indefinite equation x 2=y 5-4 has no integer solution.

The proof should be detailed, and the one who can understand is added.

We know that 3-1=2, 4-2=2, 5-3=2, so the positive integer solution of this equation has no Soling number origin group, they are.

x=n+2 y=n, where n can take all natural numbers

Therefore, there are an infinite number of positive integer solutions to the indefinite crack equation to be solved, and its solution is indefinite

Because x2-y2 (x+y)(x-y).

And because 17 is a prime number.

So x2-y 2 has no integer solution.

Proof : Suppose there is a solution, i.e., there is a positive integer x, and y makes x 2 + y 2 = 1990, then 199|x 2+y 2 if 199|x,199|y, then 199 2|x^2,199^2|y 2, so 199 2|x 2 + y 2 = 1990, contradictory! If 199 is only divisible by one of x and y, you might as well set 199|x, then 199|x 2, and 199|x^2+y^2,∴199|y^2,∴199|Y, contradiction!

If 199 is not divisible by any of x,y, then from Fermat's theorem we get x 198 1 (mod199), y 198 1 (mod199), x 198 + y 198 2 (mod199).But x 198 + y 198 = (x 2 + y 2) (x 196 - x 194y 2 + x 192y 4 -..y 196), and 199|x^2+y^2,∴199|x 198 + y 198, contradiction!

Therefore, the assumption is not true, and the equation x 2 + y 2 = 1990 has no integer solution.

xy must be odd or even, and the square of the even number must be a multiple of 4, so xy can only be two odd numbers. If this is true, then let x=2a+1, y=2b+1 have a(a+1)+b(b+1)=497 for the integer a(a+1) can only be an even number contradictory to 497, so there is no integer solution.