
> As the title asks: does there exist $N$ such that, for any prime $p$ larger than $N$, the expression $x^4 +y^4$ takes on all values in $\mathbb{Z}/p\mathbb{Z}$?
> 


I have been thinking about this problem for days, but I failed to solve it. Does anyone know whether this is true, or does anyone know any partial results about it? 


Partial results:  

If $p=4k+3$, it easily works  

if $p=4k+1$, if $g$ is a primitive root modulo $p$, and $A\_i = \left\{ g^k : k \equiv i \pmod{4} \right\}$, then at least three of $A\_i (i=0,1,2,3)$ must be expressed.



Does the expression $x^4 +y^4$ take on all values in $\mathbb{Z}/p\mathbb{Z}$?
emtom has found the right reference, but there is a more explicit result in that book (Ireland and Rosen, *A Classical Introduction to Modern Number Theory*). In fact, Theorem 5 of Chapter 8 (on page 103) directly implies that the number $N = N\_{p,\alpha}$ of solutions to $x^4+y^4=\alpha$ in $\mathbb{F}\_p$ satisfies the inequality
$$
\left| N - p \right| \leq M\_0 + M\_1 p^{1/2}
$$
for some $M\_0$ and $M\_1$ that are described explicitly in the statement of the theorem (and from that description it easily follows they can be bounded in a way that is independent of $p$ or $\alpha$).


It then automatically follows that for sufficiently large $p$, we will have $N\_{p,\alpha}>0$ for all $\alpha$. In other words, for sufficiently large $p$, the expression $x^4+y^4$ assumes all values of $\mathbb{F}\_p$ as $x$ and $y$ run through $\mathbb{F}\_p$.


Of course, this is no better (and possibly slightly worse) than what you get from the Hasse-Weil bound which Dan Loughran referred to, but at least this reference has the virtue of providing a completely elementary proof.


