If $ a^{2} + b^{2} = c^{2}$ and $c$ is even, prove that $a$ and $b$ are both even. Question:
If $ a^{2} + b^{2} = c^{2}$ and $c$ is even, prove that $a$ and $b$ are both even.
I am not quite sure how to prove this. My guess is proof by contradiction.
Assume the contrary, that is, $a^{2} + b^{2} = c^{2}$ for $c$ even and at least one of $a,b$ odd. 
 A: $c$ is even, thus $c^2=4n$ for some integer $n$.
Now if $a,b$ are both odd, then 
$$a^2+b^2=(2k+1)^2 + (2m +1)^2$$
$$=4k^2+ 4k + 1 + 4m^2 + 4m+ 1$$
$$=4n' + 2 \equiv 2 \pmod4$$
Thus $4 \nmid (a^2+b^2)$, while $4 \mid c^2$ - contradiction.
Also it is obvious that if $a$ even and $b$ odd, or $a$ odd and $b$ even; then $a^2+b^2$ is odd, while $c^2$ is even - contradiction.
Thus it has to be that both $a$, $b$ are even.
A: Since $c$ is even then $c=2k$ for some $k$. Now, suppose $a$ is odd, then $a=2m + 1$ therefore: $4m^2 + 4m + 1 +b^2 = 4k^2$. It follows $4 | 1 + b^2$ therefore $b$ is odd, $b=2n+1$ and $4 | 2 + 4n + 4n^2$, impossible.
A: Suppose $a^2+b^2=c^2$ and $c$ is even. Since  $a^2=c^2-b^2=(c+b)(c-b)$, $a$ and $b$ have the same parity. Assume $a$ is odd. Then $a^2+b^2\equiv 2(\mod4)$, whence $c^2\equiv2(\mod 4)$ which is a contradiction since $4\mid c^2$ as $2\mid c$. Hence $a$ and $b$ are even.
A: Obviuosly:
$$odd + even \ne even,$$
$$even + odd \ne even$$
Assume both odd:
$$(2m+1)^2+(2n+1)^2=(2k)^2 \Rightarrow$$
$$((2m+1)+(2n+1))^2=(2k)^2+2(2m+1)(2n+1) \Rightarrow$$
$$4(m+n+1)^2=4k^2+2(2m+1)(2n+1).$$
The last term is not divisible by $4$, hence contradiction. Both must be even.
