Determine all solutions of the congruence $y^2≡5x^3\pmod7$ in integers $x$, $y$.

Question

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 $\operatorname{ind}_35≡5\pmod 7$.

Answer 1

If $7|x\iff7|y$

Otherwise $(xy,7)=1$

Let ind$_3y=Y$ etc.

$\implies2Y\equiv5+3X\pmod6$

$\iff2(Y-1)\equiv3(X+1)$

$\implies3|(Y-1),Y=3a+1$(say)

$\implies y\equiv3^{1+3a}\equiv(-1)^a3\pmod7$

and $2|(X+1)\iff2|(X-1),X=2b+1$(say)

$\implies x\equiv3(9^b)\pmod7\equiv3\cdot2^b$

where $a,b$ are arbitrary integers

Answer 2

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.