IMC 2007 · Day 2 · P8

We claim that $29 \mid x, y, z$. Then, $x^4 + y^4 + z^4$ is clearly divisible by $29^4$.

Assume, to the contrary, that 29 does not divide all of the numbers $x, y, z$. Without loss of generality, we can suppose that $29 \nmid x$. Since the residue classes modulo 29 form a field, there is some $w \in \mathbb{Z}$ such that $xw \equiv 1 \pmod{29}$. Then, $(xw)^4 + (yw)^4 + (zw)^4$ is also divisible by 29. So we can assume that $x \equiv 1 \pmod{29}$.

Thus, we need to show that $y^4 + z^4 \equiv -1 \pmod{29}$, i.e.\ $y^4 \equiv -1 - z^4 \pmod{29}$, is impossible. There are only eight fourth powers modulo 29, $$\begin{align*} 0 &\equiv 0^4, \\ 1 &\equiv 1^4 \equiv 12^4 \equiv 17^4 \equiv 28^4 \pmod{29}, \\ 7 &\equiv 8^4 \equiv 9^4 \equiv 20^4 \equiv 21^4 \pmod{29}, \\ 16 &\equiv 2^4 \equiv 5^4 \equiv 24^4 \equiv 27^4 \pmod{29}, \\ 20 &\equiv 6^4 \equiv 14^4 \equiv 15^4 \equiv 23^4 \pmod{29}, \\ 23 &\equiv 3^4 \equiv 7^4 \equiv 22^4 \equiv 26^4 \pmod{29}, \\ 24 &\equiv 4^4 \equiv 10^4 \equiv 19^4 \equiv 25^4 \pmod{29}, \\ 25 &\equiv 11^4 \equiv 13^4 \equiv 16^4 \equiv 18^4 \pmod{29}. \end{align*}$$ The differences $-1 - z^4$ are congruent to 28, 27, 21, 12, 8, 5, 4, and 3. None of these residue classes is listed among the fourth powers.