Sums of four fourth powers

Question

Apologies in advance if this is a naive question.

If I understand correctly, it's well-known that the Fermat quartic surface

$X = \lbrace w^4 +x^4+y^4+z^4 =0 \rbrace \subset \mathbf{P}^3$

has points over every finite field $\mathbf{F}_q$ except $\mathbf{F}_5$. If I still understand correctly, one way to check this is to use the Weil bound relating the number of $\mathbf{F}_q$–points on $X$ to the number on $\mathbf{P}^2$; this shows that $X$ must have points over $\mathbf{F}_q$ for all but a small number of (small) values of $q$, and the remaining cases can be checked by hand.

The point of my question is that this proof uses the Weil bound, and I am curious if there is an elementary proof. In other words, I am looking for the most elementary proof of the following statement:

Let $p$ be a prime number different from 5. Show that there exists an $n$ such that $np$ is a sum $a^4+b^4+c^4+d^4$ of four fourth powers (where $a$, $b$, $c$, $d$ are not all divisible by $p$).

Of course, this has something to do with Waring's problem, but a quick search on that topic didn't turn up anything on this kind of variant.

Edit: Thanks to all for the very nice answers. It seems unfair to select just one, so I'll hold on to my checkmark for the time being.

Answer 1

This special case of the problem of estimating the number of ${\bf F}_p$-points on a variety is much easier than the Weil bound: the standard estimates on Gauss sums suffice. I hope that this is elementary enough. I recite at the end of this answer a derivation of a bound that generalizes to arbitrary "diagonal hypersurfaces" and reduces the present problem to an easy finite computation (through $p=73$).

With more precise information about quartic Gauss sums (somewhat less elementary, but still known to Gauss — it's in an appendix to Disquisitiones Arithmeticæ), one can obtain an exact formula for the number of rational points. Assume that $p \equiv 1 \bmod 4$, else the count is $(p+1)^2$ and is entirely elementary (we may as well count on the quadric $w^2+x^2+y^2+z^2=0$ because in this case every number mod $p$ has as many square roots as it has fourth roots). Write $p = a^2 + b^2$ with $a$ odd. Then the number of rational points is $p^2+mp+1+4a^2$, where $m=18$ or $-6$ according as $p$ is congruent to $1$ or $5 \bmod 8$. This shows that $p=5$ is the only case where there are no rational points. The formula also fits in with the fact that the diagonal Fermat quartic is a K3 surface of maximal Picard number (= Néron-Severi rank), so I'll add the "k3-surfaces" tag.

To connect the enumeration problem with Gauss sums we argue as follows. Again it is enough to consider the case $p \equiv 1 \bmod 4$, since if $p \equiv 3 \bmod 4$ the problem is equivalent to the easier task of enumerating rational points on $w^2+x^2+y^2+z^2=0$, for which the same technique works (using only quadratic Gauss sums) but is overkill.

For $n \in \bf Z$ write $e(n) = \exp(2\pi i n /p)$. Then the number of solutions of $w^4+x^4+y^4+z^4 = 0$ in ${\bf F}_p$ is $p^{-1} \sum_{k=0}^{p-1} S_k^4$ where $S_k = \sum_{t=0}^{p-1} e(kt^4)$. [Proof: expand $\sum_{k=0}^{p-1} S_k^4$ as $$ \sum_{k=0}^{p-1} \left[ \mathop{\sum\sum\sum\sum}_{w,x,y,z\phantom0\bmod \phantom0p} \phantom. e(k(w^4+x^4+y^4+z^4))\right] $$ and switch the order of summation: the sum over $k$ is $p$ if $w^4+x^4+y^4+z^4 = 0$, and zero otherwise. NB we're counting affine solutions, including $(w,x,y,z)=(0,0,0,0)$, not projective solutions.] Now $S_0 = p$. I claim that $|S_k| \leq 3\sqrt{p}$ for $k \neq 0$. It will follow that the number of solutions is within $81(p-1)p^2$ of $p^4$. In particular the number of solutions exceeds $1$ once $p \geq 81$, so the number of projective ${\bf F}_p$-points is positive. Checking the remaining cases $p=13,17,29,37,41,53,61,73$ is routine.

To prove that $|S_k| \leq 3 \sqrt{p}$ for $k \neq 0$, write $S_k = \sum_{n=0}^{p-1} c_n e(kn)$ where $c_n$ is the number of solutions of $t^4 \equiv n \bmod p$. Now $c_n = 1 + \chi(n) + \chi^2(n) + \chi^3(n)$ where $\chi$ is a Dirichlet character mod $p$ of order $4$. Thus $$ S_k(n) = \sum_{j=0}^3 \left[ \sum_{n=0}^{p-1} \phantom. \chi^j(n) \phantom. e(kn) \right]. $$ For $j=0$, the inner sum vanishes; for $j = 1,2,3$ it's a Gauss sum, which has absolute value $\sqrt p$. Hence ${}$$|S_k| \leq \sum_{j=1}^3 \sqrt p = 3 \sqrt p$ and we're done.

I'll leave the proof of the formula $p^2+mp+1+4a^2$ as an "exercise" because this answer is already rather long...

Answer 2

Expanding on Felipe Voloch's answer (and the comments which appeared while I was writing this). We want to show that $$x^4+y^4+z^4+t^4\equiv 0\pmod{p}$$ has a non-trivial solution (not all $0\pmod{p}$). I'll assume $p\equiv 1\pmod{4}. $Let $S$ be the set of fourth powers in $\mathbb F_p$ which has cardinality $\frac{p+3}{4}$ and consider the sumset $S+S+S+S/\lbrace 0\rbrace$. By Cauchy-Davenport it has size $\geq p-1$.

The only way to get equality is given by Vosper's theorem if: (1) we have $|S/\lbrace 0\rbrace|=1$ which happens only if $p=5$ or (2) The sets $S$ and $S/\lbrace 0\rbrace$ are arithmetic progressions with common difference. This means that $S=\lbrace 0,d,2d,\dots,\frac{p-1}{4}d\rbrace$ for some $d$. It is not hard to see that this implies $\frac{kd}{d}\in S$ so that in fact $S=\lbrace 0,1,\dots,\frac{p-1}{4}\rbrace$. This is obviously not possible since then $2^{p-2}$ will also be a fourth power, but $2^{p-2}\equiv \frac{p+1}{2}$ is not an element of $S$.

Answer 3

Using simple manipulations of exponential sums and the simple fact that Gauss sums modulo $p$ are of size $p^{1/2}$ one can prove the following result.

Proposition. Let $G$ be a subgroup of $\mathbb{F}_p^\times$. If $|G|>p^{\frac{k+1}{2k}}$, then every element of $\mathbb{F}_p^\times$ is a sum of $k$ elements from $G$.

Using this result one can see that for $p\geq 67$ every element of $\mathbb{F}_p^\times$ is a sum of three fourth powers of $\mathbb{F}_p^\times$, hence in particular zero is a sum of four fourth powers of $\mathbb{F}_p^\times$.

Answer 4

There are elementary ways of getting estimates on the number $N$ of elements of $\mathbb{F}_q$ which are sums of two fourth powers, e.g. Mordell's method of second moments. Once you know that something like $N > (1-\epsilon)q$, it's trivial to show that your surface has a point. Even $N > (q+1)/2$ will do, I think.