Originally Posted by

**Th3sandm4n** I searched around and couldn't find anything like this, so here's my questions:

Show that if an integer x is the sum of two squares, then $\displaystyle x \not\equiv 3 mod 4$

proof:

$\displaystyle x = a^2 + b^2$ for a,b integers

Assume $\displaystyle x \equiv 3 mod 4$. (going for a contradiction here)

$\displaystyle a^2+b^2 \equiv 3 mod 4 $ -subsitution

$\displaystyle 4 | (a^2+b^2-3)$ -def. of congruence

$\displaystyle a^2+b^2-3 = 4k$ for some k int - def. of divides

So, I have to some how show that this ISN'T true. I'm not sure if I am even going about this correctly by contradiction, but it just seems like what I would need to do.

Thanks