How to prove that $11...111$ is not the sum of two perfect squares I'm stuck with this problem: 
Show that $a=11...111$ is not the sum of two perfect squares. That is to say, there are no pair of  integers ($b$ , $c$) so that $b^2+c^2=a$. I think I am supposed to use equivalence classes in some way, but I do not know how to approach it.

Since $a$ is odd, $b$ and $c$ have different parity: $b$ is even and $c$ is odd, say.
Then $b^2+c^2\equiv 1\pmod 4$, but $a\equiv 3\pmod 4$.
In fact, no sum of two squares ends with $11$.
