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For pedagogical reasons, I got interested in the equation $y^4-x^3=a$ over $\mathbf F_p$.

To my surprise (maybe I'm naive), there is only one couple $(p,a)=(13,7)$ for which there is no solution, at least for $p\leq 2000$.

My question :

Is $(p,a)=(13,7)$ the only couple for which $y^4-x^3=a$ has no solution over $\mathbf F_p$ ?

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    $\begingroup$ Side remark: $p=13$ is a great candidate for one of these to have no solutions, since it is both $1\pmod4$ and $1\pmod3$, so that the number of 4th powers and 3rd powers is as small as possible. $\endgroup$
    – Greg Martin
    Commented Jan 17, 2019 at 18:55

1 Answer 1

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The curve $C:y^4-x^3z=az^4$ is nonsingular over $\mathbb F_p$ for $p\ge5$ and $a\ne0$. It has genus $3$. So Weil's theorem says that $$ \bigl| \#C(\mathbb F_p) - p - 1 \bigr| \le 6\sqrt{p}. $$ There is only one point with $z=0$, namely $[1,0,0]$, so you'll always get solutions to your original equeation provided $p>6\sqrt{p}$, i.e., provided $p>36$. For $p<36$, it is a finite calculation to check, which you have already done.

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  • $\begingroup$ Very nice, thanks a lot $\endgroup$
    – few_reps
    Commented Jan 17, 2019 at 14:09

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