## RE Perfect square problem

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• A + x^2 = y^2 A = y^2 - x^2 A = (y+x)(y-x) It depends on the number of divisions of A in two divisors. There is a finite number of solutions to the equation.
Message 1 of 1 , Oct 21, 2004
A + x^2 = y^2

A = y^2 - x^2

A = (y+x)(y-x)

It depends on the number of divisions of A in two divisors. There is a
finite number of solutions to the equation. You can't find an y for every x.
You must equal y+x to the biggest divisor F and y-x to the smallest f, where
F*f = A. Then y = (F+f) / 2 and x = (F-f) / 2. You will get a solution only
for those divisors whose sum is even, so the numbers n = 2*k where MCD(k,2)
= 1 don't have a solution.

For example:

A = 12 --> A = 12 * 1 = 6 * 2 = 4 * 3, only 6 + 2 is even.

x+y = 6
y-x = 2

x = 2, y = 4

12 + 2^2 = 4^2 --> 12 + 4 = 16 Ok.

I hope it's useful.

Jose Brox

----- Original Message -----
From: "Guna" <gunaseelapandian@...>
Sent: Thursday, October 21, 2004 1:42 PM

>
> Is there a way mathematically to find out two numbers such that one is a
square (I mean whole) when a known number is added with the second's square
?
>
> for example let A be a number that is known
>
> now we add (x)2 to A and the result is a perfect square y such that the
square root of y is lets say r that does not have a decimal part.
>
> I am not a mathematician but am working on primes for almost 3 months now.
This is really fascinating.
>
> Guna
>
>
>
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