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!
 A: 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 $p \equiv 1 \pmod{4}$.
A: As a first step, it is easy to see that the multiple $6p^2$ can be written as the sum of $12$ fourth powers. The reason is, that every integer of the form $6x^2$ can be written this way, because $x=a^2+b^2+c^2+d^2$ is the sum of four squares, and 
\begin{align*}
6x^2 & =6(a^2+b^2+c^2+d^2)^2 \\
     & =(a+b)^4+(a-b)^4+(a+c)^4+(a-c)^4+(a+d)^2+(a-d)^4 \\
     & +(b+c)^4+(b-c)^4+(b+d)^4+(b-d)^4+(c+d)^4+(c-d)^4.
\end{align*}
To show that $rp=a^4+b^4+c^4+d^4+e^4$ for some $r<p^3/4$ one could perhaps use Minkowski's lattice theorem (which is used to prove that every integer is a sum of four squares etc.).
