Sum of powers divisible by 53 
Show that $1^{26} + 2^{26} + 3^{26} + \cdots + 26^{26} $ is divisible by $53$.

Inspired by another more basic question asked here... I'm interested to see what elegant proofs you come up with (and mine is also given below in an answer).

$53$ is a prime of the form $1 \bmod 4$. Consider a primitive root $g$ - this has quadratic residues at even powers and non-residues at odd powers. So we know that $-1$ is a quadratic residue, and thus for any $x$ a quadratic residue, $-x$ is also. Also since exactly half the totients of $53$ are quadratic residues, we must have that half the numbers in $\{1,2,\ldots,26\}$ are quadratic residues and half are quadratic non-residues. For a quadratic residue $q$, $q^{26}\equiv 1 \bmod 53$ and for a quadratic non-residue $n$, $n^{26}\equiv -1 \bmod 53$. 
Thus we have $1^{26} + 2^{26} + 3^{26} + \cdots + 26^{26} \equiv 13\cdot1 + 13\cdot -1 \equiv 0 \bmod 53$
