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 $\mathbf F\_p$ ?



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


