<s>[INST] Determine all solutions of the congruence $y^2≡5x^3\pmod7$ in integers $x$, $y$. Determine all solutions of the congruence $y^2≡5x^3\pmod7$ in integers $x$, $y$.
I learned about primitive roots and the theory of indices.
By trial, I check that $3$ is a primitive root of $7$ and $\operatorname{ind}_35≡5\pmod 7$.
 [/INST] With $p=7$, you have that $6|p-1$, which means there are only a limited number of squares and cubes mod $7$. Whatever $y$ is, $y^2$ can only be $0$, $1$, $4$, or $2$. 
And $x^3$ can only be congruent to $0$, $1$, or $6$. So $5x^3$ can only be $0$, $5$, or $2$.
So with the given congruence, for the left side to match the right side, either both sides are $0$ or both sides are $2$.
If both sides are $0$, this implies $x\equiv y\equiv 0$.
If both sides are $2$, then $y$ could be either square root of $2$ (which are $3$ and $4$) and $x$ could be any cube root of $2/5\equiv6$ (which are $3$, $5$, and $6$.)
All together that makes $2\cdot3+1=7$ solution pairs.
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