Find all solutions to $y^2 \equiv 5x^3 \pmod {7}$ 
Find all solutions to $y^2 \equiv 5x^3 \pmod {7}$

So basically one can evaluate all $a^2, 5b^3 \pmod {7}$ and look for all $(a,b) \in \mathbb Z_7\times \mathbb Z _7$ such that $a^2 \equiv 5b^3 \pmod 7$
Two questions: 


*

*Is that the "best" way doing it?

*If I counted correctly there are exactly $7$ solutions. Is that a coincidence it equals to the modulus? 


Thanks!

If  $7\mid x\iff7|y$
else $7\nmid xy$
In that case as $\phi(7)=6, x^3\equiv\pm1\pmod7$
If $x^3\equiv1,y^2\equiv5$
Now as $y\equiv\pm1,\pm2,\pm3\pmod7, y^2\equiv1,4,9\equiv2$
Hence, $y^2\not\equiv5$
If $x^3\equiv-1,y^2\equiv-5\equiv2\equiv3^2\implies y\equiv\pm3$
Now $x^3\equiv1\implies x\equiv3,5\equiv-2,6\equiv-1\pmod7$
