Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even
My Attempt, Case by case analysis:
Case 1: a is odd, b is odd. From the first equation,
$odd^2 + odd^2 = c^2$
$odd + odd = c^2 \implies c^2 = even$
Squaring a number does not change its congruence mod 2.
Therefore c is even
$ a + b + c = odd + odd + even = even$
Case 2: a is even, b is even. Similar to above
$even^2 + even^2 = c^2 \implies c$ is even
$a + b + c = even + even + even = even$
Case 3:
One of a and b is odd, the other is even
Without loss of generality, we label a as odd, and b as even
$odd^2 + even^2 = c^2 \implies odd + even = c^2 = odd$
Therefore c is odd
$a + b + c = odd + even + odd = even$
We have exhausted every possible case, and each shows $a + b + c$ is even. QED
Follow Up:
Is there a proof that doesn't rely on case by case analysis?
Can the above be written in a simpler way?
Best Answer
Note that $x^2\equiv x\pmod 2$ and thus $a^2+b^2=c^2$ implies $$a+b+c\equiv a^2+b^2+c^2\equiv 2c^2\equiv 0\pmod 2$$