If $ a^{2} + b^{2} = c^{2}$ and $c$ is even, prove that $a$ and $b$ are both even.

Question

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.

Answer 1

$c$ 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.

Answer 2

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.

Answer 3

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.

Answer 4

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.