TITLE: Does the expression $x^4 +y^4$ take on all values in $\mathbb{Z}/p\mathbb{Z}$?
QUESTION [9 upvotes]: 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.

REPLY [12 votes]: Expanding on a comment, the curve $X^4+Y^4=aZ^4$ (for $a\ne0$) has genus $3$. So the Hasse-Weil bound says
$$ N_p(a) := \#\bigl\{ [X,Y,Z]\in\mathbb P^2(\mathbb F_p) : X^4+Y^4=aZ^4 \bigr\} $$
satisfies
$$ \bigl| N_p(a) - p - 1 \bigr| \le 2g\sqrt{p} = 6\sqrt{p}. $$
Thus
$$ N_p(a) \ge p + 1 - 6\sqrt{p}. $$
There are at most $4$ points with $Z=0$, so
$$
\#\bigl\{ (X,Y) \in\mathbb A^2(\mathbb F_p) : X^4+Y^4=a \bigr\} \ge p-3-6\sqrt{p}.
$$
So you'll always have a solution provided $p\ge3+6\sqrt{p}$, which means that there is always a solution provided $p\ge43$.