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.
Best Answer
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.