# not sum of two squares

Let $m=r\cdot s$ where $m,r,s$ are natural numbers.

Let us ascribe a number as re-presentable if it can be expressed as the sum of two squares in integers.

We can prove, if any two of $m,r$ and $s$ are re-presentable, so will be the other.

How to prove if at least one of $r,s$ is not re-presentable, $m$ will not be re-presentable.

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If you know the theorem that tells you exactly which numbers are representable as a sum of two squares, this is easy. If you know about unique factorization in the Gaussian integer, this is easy. Can you use either of those two approaches? –  Gerry Myerson Oct 17 '12 at 6:30
@GerryMyerson, thanks for your instructive input. I observe, if both $r,s$ contain odd powers of the same prime$\equiv -1 \pmod 4$, $m$ becomes re-presentable, even though $r,s$ are not. $\prod p_i$ must be $\equiv 1,2\pmod 4$ to be re-presentable can only be product of such prime factors, right? –  lab bhattacharjee Oct 17 '12 at 6:50
Yes, a number is a sum of two squares if and only if there is no 3-mod-4 prime dividing the number to an odd power. –  Gerry Myerson Oct 17 '12 at 21:40