At least one member of a pythagorean triple is even 
I am required to prove that if $a$, $b$, and $c$ are integers such that $a^2 + b^2 = c^2$, then at least one of $a$ and $b$ is even. A hint has been provided to use contradiction. 

I reasoned as follows, but drew a blank in no time:
Let us instead assume that both $a$ and $b$ are odd. This means that $a^2$ and $b^2$ are odd, which means that $c^2$ is even. Thus, $c$ is even. But this is not a contradiction -- it's the sum of two odd numbers, after all.
 A: Suppose that $a$ and $b$ are both odd, and proceed as you did to conclude that $c$ is even. As $a$ is odd, we may write $a = 2k + 1$ for some $k$; similarly, $b = 2m + 1$. As $c$ is even, write $c = 2n$. This leads to
$$(2k + 1)^2 + (2m + 1)^2 = (2n)^2$$
or upon expanding and regrouping,
$$4(k^2 + k + m^2 + m) + 2 = 4n^2$$
Now the right side is divisible by 4, as is the first term on the left - but $2$ is not divisible by $4$. Do you now see how to derive a contradiction?
A: Let us try to write this using congruences, which is a very useful and compact notation.
We know that a square can only have remainder $0$ or $1$ modulo $4$. I.e., for any integer $x$ one of these two congruences must be true: $x^2\equiv 0 \pmod 4$ or $x^2\equiv1 \pmod4$. The first case happens if $x$ is even, the second case if $x$ is odd. (Since $(2k+1)^2=4(k^2+k)+1$ and $(2k)^2=4k^2$.)
So we have two possibilities for the remainder of $a^2$ and two possibilities for the remainder of $b^2$ modulo $4$.
$$
a^2 \equiv 0 \pmod4, b^2 \equiv 0 \pmod4 \Rightarrow c^2\equiv 0+0=0 \pmod4\\
a^2 \equiv 1 \pmod4, b^2 \equiv 0 \pmod4 \Rightarrow c^2\equiv 1+0=1 \pmod4\\
a^2 \equiv 0 \pmod4, b^2 \equiv 1 \pmod4 \Rightarrow c^2\equiv 0+1=0 \pmod4\\
a^2 \equiv 1 \pmod4, b^2 \equiv 1 \pmod4 \Rightarrow c^2\equiv 1+1=2 \pmod4
$$
We see that in the last case we would have $c^2\equiv 2\pmod4$, which is not possible.
So only the first three cases can really occur. In the other words, either all of the numbers $a$, $b$, $c$ are even, or exactly one of the is even and the remaining two numbers are odd.
A: Many formulas generate valid Pythagorean triples that include all primitives, among them are Euclid's:
$$A=m^2-n^2\qquad B=2mn\qquad C=m^2+n^2$$
and one I developed where $GCD(A,B,C)$ is always an odd square
\begin{equation}
  A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2
\end{equation}
By inspection, we can see that  side-B is always even, no matter what the multiplier so one side is always even.
