Is there any way to get two factors of a number whose sum of the squares is a perfect square. As an example

$19354423920$ is a number. which has $4262$, $4541160$ as factors ($19354423920 = 4262 * 4541160$).

The sum of the squares of the factors will then be a perfect square like this $4262^2 + 4541160^2 = 4541162^2$.

How does one find out this kind of factors from a number?

  • $\begingroup$ To be clear, you're looking for numbers that have factors that are part of a Pythagorean triple? $\endgroup$ – Arthur Jan 22 '16 at 21:18
  • $\begingroup$ exactly. like 'n' is a number who has 2 factors 'a' and 'b' where a^2+b^2 = c^2 $\endgroup$ – Bashar Jan 22 '16 at 22:57

We know that Pythagorean triples have the form $a(m^2-n^2), 2amn, a(m^2+n^2)$ where $m$ and $n$ are coprime and of opposite parity. Any number of the form $2a^2mn(m^2-n^2)=2a^2mn(m+n)(m-n)$ can be factored to give two legs of a Pythagorean triangle. It is easier to generate such numbers than to find such a factorization-just choose $a,m,n$ and compute it. Most numbers will not have such a factorization. Even if it has one, if it has lots of prime factors it may not be easy to find the proper factorization.

  • $\begingroup$ Thanks for the explanation $\endgroup$ – Bashar Jan 23 '16 at 0:14

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