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.

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    $\begingroup$ I'm not sure whether you're using $\mathbb{Z}_p$ to denote the ring of $p$-adics or for $\mathbf{Z}/p\mathbf{Z}$. $\endgroup$ – YCor Apr 1 '20 at 13:32
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    $\begingroup$ @YCor I'm not sure that matters, in the light of Hensel's lemma...by which I suspect that if the statement holds for $\mathbb{Z}/p\mathbb{Z}$, then it holds for $\mathbb{Z}_p$. $\endgroup$ – Franka Waaldijk Apr 1 '20 at 13:39
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    $\begingroup$ @FrankaWaaldijk sure but it would matter in the way to formulate an answer. $\endgroup$ – YCor Apr 1 '20 at 13:43
  • $\begingroup$ Ok, I get you now :-) $\endgroup$ – Franka Waaldijk Apr 1 '20 at 13:43
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    $\begingroup$ Actually it is quite important whether $\mathbb{Z}_p$ means the $p$-adic integers or the finite field. In the latter case the answer is yes, but not in the former case. The problem is that $x^4 + y^4 = p$ has a solution in the $p$-adics if and only if $p$ is split in the $8$th cyclotomic field (i.e. $p \equiv 1 \bmod 8$). The easiest way I know how to prove it over a finite field is via the Hasse-Weil theorem, but probably there is an elementary approach. $\endgroup$ – Daniel Loughran Apr 1 '20 at 14:34

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.


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 $p\ge43$.

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    $\begingroup$ By adding a numerical computation, it seems to hold for all $p \geq 31$. For $p=29$, we get e.g. that there are no $x,y \in \mathbb{F}_{29}$ such that $x^4+y^4=4$. $\endgroup$ – RP_ Apr 1 '20 at 18:42
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    $\begingroup$ @RP That makes sense, I took the worst possible case that there are 4 $\mathbb F_p$ rational points at infinity, but for many values of $p$, there will be less. It ultimately depends on whether $-1$ is a square root, and whether it is a fourth root. $\endgroup$ – Joe Silverman Apr 1 '20 at 18:57

Not exactly an answer, but in exercise $18$ in page $106$ of Ireland and Rosen's A Classical Introduction to Modern Number theory it states: Let $p\equiv 1\mod 4$ and let $p=A^2+B^2$ where we fix $A$ by requiring that $A\equiv 1\mod 4$. Then $N=\#\{(x,y)\in\mathbb{F}_p^2\mid x^4+y^4=1\}$ satisfies $N=p-3-6A$ if $p\equiv1\mod 8$ and $N=p+1+2A$ if $p\equiv5\mod 8$.


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