Prove that the sum of pythagorean triples is always even

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

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

Hint

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

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