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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.

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    $\begingroup$ Do you mean for n to be relatively prime to p? $\endgroup$ – Gjergji Zaimi Sep 8 '11 at 19:11
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    $\begingroup$ Quick thought: The hardest case is when $p \equiv 1\bmod 4$, since otherwise every square mod $p$ is a fourth power and there are lots of those. If $p \equiv 1\pmod{4}$, let $A$ be the subset of fourth powers mod $p$ ($0$ included) and $B$ the subset of nonzero fourth powers. It looks like successive application of the Cauchy-Davenport inequality shows that, the sum set $A+A+A+B$ has size at least $p-1$, so if $0$ is missing, it's the only missing element. Moreover, the only way $0$ could be missing is if equality holds at every stage. $\endgroup$ – Anonymous Sep 8 '11 at 19:27
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    $\begingroup$ Gjergji: I certainly mean something, namely that a,b,c,d shouldn't all be divisible by p. (Relative primality seems stronger than that, but maybe I am being dumb...) $\endgroup$ – user5117 Sep 8 '11 at 19:40
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    $\begingroup$ It's also worth noting that in cases like this, the estimates from Weil's Riemann Hypothesis can be proved by using the classical theory of Gauss and Jacobi sums. See chapter 8, section 7 of Ireland and Rosen's beautiful book. $\endgroup$ – Anonymous Sep 8 '11 at 19:42
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    $\begingroup$ Cannot resist: among many similar items, I have a conjecture that $$ (w_1^4 + w_2^4 + w_3^4 + w_4^4) + 5 (x_1^4 + x_2^4 + x_3^4 + x_4^4) + 25 (y_1^4 + y_2^4 + y_3^4 + y_4^4) + 125 (z_1^4 + z_2^4 + z_3^4 + z_4^4) $$ integrally represents all positive integers. $\endgroup$ – Will Jagy Sep 8 '11 at 20:06
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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...

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  • $\begingroup$ In fact my response was based on a similar (but slightly more general) argument. $\endgroup$ – GH from MO Sep 9 '11 at 1:07
  • $\begingroup$ @GH: Yes, so I surmised, and probably the same for the comment on the original question by "anonymous" that mentioned Ireland-Rosen. I thought it would still be interesting to see the explicit answer in terms of $p=a^2+b^2$ for the diagonal quartic surface. $\endgroup$ – Noam D. Elkies Sep 9 '11 at 1:15
  • $\begingroup$ @Noam: Definitely your explication is interesting, and surely details are always useful. I was lazy to type in my proof which is standard anyways (although the statement is worthwhile I think). $\endgroup$ – GH from MO Sep 9 '11 at 1:46
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    $\begingroup$ If you want to shave off a few of the special cases, note that, if $p \equiv 1 \mod 3$, then $\mathbb{F}_p$ has a primitive cube root of unity $\omega$, and $(1, \omega, \omega^2, 0)$ is on the surface. So we only need to check $29$ and $53$ by hand. $\endgroup$ – David E Speyer Sep 9 '11 at 13:31
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    $\begingroup$ @David S.: I guess you've already dealt with 17 and 41 (which are congruent to 1 mod 8 but not mod 3) using $(1,x,1,x)$ where $x$ is an 8th root of 1. $\endgroup$ – Noam D. Elkies Sep 9 '11 at 14:03
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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 $S$.

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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 $\mathbb{F}_p^\times$.

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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.

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