Find all the solutions of $y^2 \equiv 5x^3 \pmod 7$ I need to find all the solutions of $y^2 \equiv 5x^3 \pmod 7$.
I managed to solve by trying one-by-one, but I guess there is some other way to solve this.
 A: Any solution $(x,y)$ is a point in the plane $\mathbb{Z}_7 \times \mathbb{Z}_7$. Any line $L \subset \mathbb{Z}_7 \times \mathbb{Z}_7$ through $(0,0)$ contains $3$ solutions since $y^2 - 5x^3 = 0$ has degree $3$. But the $(0,0)$ solution counts by $2$. So to get all the solutions it is enough to parameterize lines through the origen an look in each of them the third solution. Lines which are not parallel to the $y$ axis are parameterized by $t \in \mathbb{Z}_7$ as follows $$L_t := \{ (s,st) : s \in \mathbb{Z}_7\}$$
A point in $L_t$ is a solution if $5s^3 = (st)^2$, so $$s = 3t^2 $$
since we are looking solution in $L_t$ different that $(0,0)$. Then the point $(3t^2,3t^3)$ is the unique solution in $L_t$ different from $(0,0)$.
Given values $t=1,2,3,4,5,6$ we get the solutions $$(3,3),(5,3),(6,4),(6,3),(5,4),(3,4) \, .$$ These together with $(0,0)$ are all the solutions since on the $y$ axis the only solution is $(0,0)$.
A: Let $S$ be the set of all possible remainders of $y^2 \mod 7$ and $T$ be the set of all possible remainders of $5x^3 \mod 7$, 
If we try $0$ to $6$ we shall see $S=\{0,1,2,3,4\}$ and $T=\{0,2,5\}$, thus the intersection of these two sets is $\{0,2\}.$
In the conclusion, the set of ordered pairs for desired $x$, $y$ is following:
$\{(x,y)|5x^3\equiv y^2\equiv2\mod 7\}\cup\{(x,y)|5x^3\equiv y^2\equiv0\mod7\}$
Again by checking which numbers from $0$ to $6$ have the properties, we will see the answer is:
$\{7k+3,7k+5,7k+6\}\times\{7k+3, 7k+4\}\cup\{7k\}\times\{7k\}$ $(k\in\mathbb Z)$
