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.

Let $a=2n+1$, $b=2m+1$. Then $a^2 + b^2=4n^2 + 4n +4m^2 +4m+2$. This is divisible by $2$, a prime number, but not by $4=2^2$. Hence it cannot be the square of an integer.
