Prove that the sum of pythagorean triples is always even

Solution 1:

Note that $x^2\equiv x\pmod 2$ and thus $a^2+b^2=c^2$ implies $$a+b+c\equiv a^2+b^2+c^2\equiv 2c^2\equiv 0\pmod 2$$

Solution 2:

Also, Pythagorean triples have a well defined structure: $$a=k(m^{2}-n^{2}),\ \,b=k(2mn),\ \,c=k(m^{2}+n^{2})$$ and $$a+b+c=2k(mn+m^2)$$

Solution 3:

Hint

Write $a+b+c=k$, so

$$a^2+b^2=(a+b)^2-2ab= (k-c)^2-2ab=c^2 → k^2-2(kc+ab)=0→k^2=2(kc+ab)$$

Solution 4:

Notice that $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=2c^2+2(ab+bc+ac)$, so the square of $a+b+c$ is even and thus $a+b+c$ is also even.

Solution 5:

$c^2 = a^2 + b^2 = (a+b)^2 - 2ab$.

$2ab = (a+b)^2 - c^2$

$2ab = (a+b+c)(a+b-c)$

Let $n = a+b+c$, and the above becomes:

$2ab = n(n-2c)$

So the right-hand side must be even, but since $n-2c$ is odd when $n$ is odd, $n$ must be even.