2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.