Given $a^2+b^3=c^4+d^5$, prove a+b+c+d is even (for natural numbers)

Question

All I can say is that either all numbers are odd, all numbers are even, or exactly two of them are even. How will a proof of that go?

Answer 1

If you know modular arithmetic,

$n^k\equiv n\bmod 2,$ so $a^2+b^3=c^4+d^5$ means

$a+b+c+d\equiv a+b-c-d\equiv a^2+b^3-(c^4+d^5)\equiv0\mod 2.$