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Let $p$ be a prime. Prove that the number of ordered quadruples $(x,y,z,t)$ with $x^3 + y^3 + z^3 + t^3 = 0 \pmod p$ is $p^3 + 6p^2 - 6p$ if $p\equiv 1 \pmod 3$ and $p^3$ otherwise.

I can deal only with the second case. By considering a primitive root one can show that for any $b$ there is a unique $x$ with $x^3 \equiv b \pmod p$. Hence in $x^3 + y^3 + z^3 + t^3 = 0 \pmod p$ we can choose $y, z, t$ freely and from them $x$ is uniquely determined - so in total there are $p^3$ solutions.

What about $p\equiv 1 \pmod 3$? Would perhaps some tricks with Gauss sums help? (If possible let's avoid them, though.) Any help appreciated!

Update: Algebraic Geometry answers are appreciated, but can somebody give a purely elementary solution?

DesmondMiles
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1 Answers1

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If you remove the solution $[0,0,0,0]$, and then say two solutions are equivalent if they are the same up to a scalar in $\mathbf{F}^{\times}_p$, then you are just counting the number of points mod-$p$ on the corresponding projective variety:

$$X: x^3+y^3+z^3+t^3=0$$

in $\mathbf{P}^3$. This is a well-known variety called the Fermat cubic, a (rational) cubic surface which isomorphic to $\mathbf{P}^2$ blown up at $6$ points defined over $\mathbf{Q}(\sqrt{-3})$. If $p \equiv 2 \bmod 3$, these base points are not rational, and so you just get $|\mathbf{P}^2(\mathbf{F}_p)| = p^2+p+1$. If $p \equiv 1 \bmod 3$, then each of these points give an exceptional $\mathbf{P}^1$ with $p+1$ points (or $p$ more points), and thus there are

$$(p^2 + p + 1) + 6p = p^2 + 7p + 1$$

such points. This gives your formulas, i.e.

$$(p-1)(p^2+p+1) + 1 = p^3,$$ and $$(p-1)(p^2+7p+1) + 1 = p^3 + 6p^2 - 6p$$ respectively.

For explicit parameterizations of $X$ and the realization of $X$ as $\mathbf{P}^2$ blown up at $6$ explicit points, see here: http://people.math.harvard.edu/~elkies/4cubes.html

user902362
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