Is "$a^p + b^p = c^p + d^p \Rightarrow (a+b) \equiv (c+d) \mod p$" true?

Just found another nice proof of this fact (for n = 5, but can "easily" be generalised to all $p$): by the binomial theorem, $$(a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5$$ which can be rearranged to show that $$a^5+b^5 = (a+b)^5 - 5(a^4b + 2a^2b^3 + 2a^3b^2 + ab^4)$$ $$\Rightarrow a^5 + b^5 \equiv (a+b)^5 \mod 5$$ $$\therefore a^5 + b^5 \equiv c^5 + b^5 \mod 5 \Rightarrow (a+b)^5 \equiv (c+d)^5 \mod 5$$

This is perfectly fine, the only detail is that you absolutely don't need the four integers to be distinct, they could be anything.