Prove, that every $p$ prime has a multiple, which is smaller than $\frac{p^4}{4}$, it can be written down as the sum of five integer's fourth power. I have a really interesting (and hard) number theory task:
Prove, that every $p$ prime has a multiple(not $0$), which is smaller than $\frac{p^4}{4}$, and it can be written down as the sum of five integer's fourth power.
I have tried it for smaller $p$ numbers and I couldn't even find an example for it. Is this statement true or not, if so, how should I prove it? :)
Thanks!

If $p \equiv 3 \pmod{4}$, there are $\frac{p-1}{2}$ quartic residues, and by the pigeonhole principle there is at least one $k$ such that both $k$ and $p-k$ are quartic residues (except for the edge case that all $0 < k < \frac{p-1}{2}$ are quartic residues and all other values are not, that possibility can be handled similarly).  Setting $k \equiv a^4 \pmod{p}$ and $p-k \equiv b^4 \pmod{p}$, we get $n = a^4 + a^4 + b^4 + b^4 + 0^4 \equiv 0 \pmod{p}$.  WLOG $a < \frac{p}{2}$ and $b < \frac{p}{2}$, so $a^4 < \frac{p^4}{16}$ and $b^4 < \frac{p^4}{16}$, and $n \lt \frac{p^4}{4}$ as desired.
Something like that might also work for $p \equiv 1 \pmod{4}$.
