Prove that the sum of the squares of two odd integers cannot be the square of an integer. Prove that the sum of the squares of two odd integers cannot be the square of an integer.
My method:
Assume to the contrary that the sum of the squares of two odd integers can be the square of an integer. Suppose that $x, y, z \in \mathbb{Z}$ such that $x^2 + y^2 = z^2$, and $x$ and $y$ are odd. Let $x = 2m + 1$ and $y = 2n + 1$. Hence, $x^2 + y^2$ = $(2m + 1)^2 + (2n + 1)^2$
$$= 4m^2 + 4m + 1 + 4n^2 + 4n + 1$$
$$= 4(m^2 + n^2) + 4(m + n) + 2$$
$$= 2[2(m^2 + n^2) + 2(m + n) + 1]$$
Since $2(m^2 + n^2) + 2(m + n) + 1$ is odd it shows that the sum of the squares of two odd integers cannot be the square of an integer.
This is what I have so far but I think it needs some work.

You had a great start but you should not make this last factorisation.
$$4(m^2 + n^2) + 4(m + n) + 2=4(m^2+n^2+m+n)+2\equiv 2 \pmod 4$$
But a square can't be equal to $2 \pmod 4$?
