# Universality of $y^4-x^3$ mod $p$

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 $$\mathbf F_p$$ ?

• 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. Jan 17 '19 at 18:55

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.