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.
 A: 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.
A: 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 $2 \pmod 4$?
A: Here's a quick method, not unrelated to your approach or to the other answers here.
The squares mod $4$ are $0$ and $1$ (can be verified easily by checking all four). Odd numbers are congruent to $1$ or $3$ mod $4$ and these each have square congruent to $1$ mod $4$. Hence the sum of two odd squares is congruent to $2$ mod $4$ which isn't a square.
A: just think base2:
a and b, being odd numbers will look like $$ xxxx1_2$$
square of a and b will be odd and also look  like $$ xxxx1_2 $$
sum of $$xxxx1_2 + xxxx1_2 = xxx10_2 $$
square of even number $$xxxx0_2$$ will look like $$xxx00_2$$
so digits on position '2' will be always 0 for square of even number but for sum of squares of odd numbers will be 1:
$$ xxxx10_2 \neq xxx00_2 $$
