Let $p>3$ be a prime number. The natural numbers $a,b,c, d$ are such that $a+b+c+d$ and $a^3+b^3+c^3+d^3$ are divisible by $p$. Prove that for all odd $n$, $a^n+b^n+c^n+d^n$ is divisible by $p$.
If $p|a+b$ then $p|c+d \to p|a^n+b^n$ and $p|c^n+d^n\to p|a^n+b^n+c^n+d^n$ for odd $n$
Let $p\not | (a+b)$
$p|a^3+b^3+c^3+d^3\to p|(a+b)^3+(c+d)^3-3ab(a+b)-3cd(c+d) \to p| (a+b+c+d) ( (a+b)^2+(c+d)^2-(a+b)(c+d))-3 (ab(a+b)+cd(c+d)) \to p | ab(a+b)+cd(c+d)$
$p|ab(a+b)+cd(c+d)\to p|ab(a+b)+cd(-(a+b)) \to p| (a+b)(ab-cd) \to p|ab-cd \to p|ab-c(-a-b-c) \to p| c^2-ac-bc+ab \to p|(c+a)(c+b)$
Let $p|a+c$ then $p|b+d$ and so $p| a^n+c^n+b^n+d^n$ for odd $n$