# Diophantine equation: $x^2-y^2=z$ for a fixed integer $z$

Is there any relatively efficient way to calculate integer solutions $x$ and $y$ to the equation $x^2 - y^2 = z$ for a fixed integer $z$?

May or may not be useful: $z$ is an odd composite number

Thanks

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## migrated from mathoverflow.netJul 2 '13 at 1:49

This question came from our site for professional mathematicians.

Why not just ask if there's an efficient way to factor integers? – Steven Landsburg Jul 2 '13 at 1:23

Write $z$ as $z=ab$. Hence, one solution for $x^2-y^2 = ab$ is $(x+y) = a \text{ and } x-y =b$.

I am sure you can take it from here.

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We do not have uniqueness in the case of two primes, since the trivial factorization gives a solution. – André Nicolas Jul 2 '13 at 1:55

HINT:

If $z$ is odd, $z\pm 1$ i s even $\implies \frac{z\pm1}2$ is integer

$$z\cdot1=\left(\frac{z+1}2\right)^2-\left(\frac{z-1}2\right)^2$$

If $z$ composite odd $=m\cdot n$(say), $m\pm n$ is even

$$m\cdot n=\left(\frac{m+n}2\right)^2-\left(\frac{m-n}2\right)^2$$

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