For which primes $p$ do there exist $x,y\in\mathbb{Z}_p$ such that $3y^2=4x^3-10?$
This is a question in an elliptic curves course so I assume we want to transform this into an elliptic curve! My idea is that we want to birationally transform this curve into one with a Weirstrass equation, i.e. $y^2=x^3+Ax+B$.
To get rid of the coefficient of $3$ before the $y^2$ term, we can make the mapping $(x,y)\mapsto (x,y/\sqrt{3})$ which gives $Y^2=4x^3-10$. Now we want to get rid of the coefficient before $x^3$ so we can make the overall transformation $(x,y)\mapsto\left(\frac{x}{\sqrt[3]{4}},\frac{y}{\sqrt{3}}\right)$. This will then give \begin{equation}Y^2=X^3-10.\end{equation}
So we have now recast the problem into finding the primes $p$ such that there are $X,Y\in\mathbb{Z}_p$ with $Y^2=X^3-10$. However, it is not clear to me that the transformation I made is actually going to be useful, nor can I figure out how, even if it is a good idea, I can then go on to solve the problem.
A naive idea would be to look at the constant term $10$ and consider its factors; say, for instance, we work modulo $2$, then we require $Y^2\equiv X^3\pmod 2$. And we could then check case by case; same as for $p=5$. But this doesn't seem like a particularly useful idea.
This is too loong for a comment, so it became an answer. To bring the given equation in the short Weierstraß form over the rationals, we can follow the steps:
Substitute now the number under the square on the L.H.S. by $Y$, the number under the cube on the R.H.S. by $X$, so we obtain an equation of an equivalent curve: $$ (E)\ :\qquad Y^2 = X^3 - 4320\ . $$ Which is showing only the affine piece $(X,Y)$ of points $[X:Y:Z]$ in the $2D$ projective space, where $N\ne 0$. Note that for a fixed prime $p$ a projective point $[X:Y:Z]$ can be rearranged up to a non-zero rational scalar, so that its components have no $p$-divisible denominators, and that modulo $p$ at least one component is not zero. Take such a representative when passing from $\Bbb Q$ or $\Bbb Q_p$ to $\Bbb Z/p=\Bbb F_p$. We obtain a map $\psi=\psi_p$ from points in characteristic zero to points in characteristic $p$.
Note that the operations use the primes $p=2,3$, so points in $\Bbb Z_p$ before and after the transformation, seen in both directions, may leave $\Bbb Z_p$. For these primes, we need a special analysis afterwards.
So we have obtained the form $Y^2 = X^3 +aX+b$ with $a=0$, $b=-4320$. Its discriminant is: $$ \Delta = -16(4a^3 + 27b^2) =-16(0 + 27(-2^5 \cdot 3^3 \cdot 5)^2) =-2^4\cdot 3^3\cdot(-2^5 \cdot 3^3 \cdot 5)^2 =-2^{14}\cdot 3^9\cdot 5^2\ . $$ So the bad primes are $2,3,5$.
Let now $p$ be a prime. The above equation $(E)$ lives over $\Bbb Z$, so we can consider it modulo $p$, and obtain an equation $(\bar E)$, or $E_{\Bbb F_p}$ over the field $\Bbb F_p$ with $p$ elements. Here, a bar or the subscript $\Bbb F_p$ denotes the passage from $\Bbb Z$ to $\Bbb Z/p=\Bbb F_p$. The equation $(\bar E)$ is again defining an elliptic curve, iff the discriminant $\bar\Delta \ne 0$ in $\Bbb F_p$. So for all $p\ne 2,3,5$ we obtain an elliptic curve. There are no singular points for such $p$-values.
Let $E_{\text{ns}}(\Bbb F_p)$ be the set of all $\Bbb F_p$-points which are non-singular. The elliptic curve $E$ can be also seen over $\Bbb Q_p$, and there is a morphism from $E_0(\Bbb Q_p)$, be the $\Bbb Q_p$(-rational)-points that map to non singular points via $\psi=\psi_p$. Its the set of points with good reduction modulo $p$. (The index $0$ "depends on $p" so far.)
Then this map $$ E_0(\Bbb Q_p)\overset{\psi}\longrightarrow\bar E_{\text{ns}}(\Bbb F_p) $$ is surjective.
Let $p>5$ be a prime. Since every non-singular elliptic curve has rational points, and we also know lower and upper bounds for their number, the lower Hasse bound being $(\sqrt p-1)^2$, the set $E_0(\Bbb Q_p)$ is not empty, and there is at least one point $\ne \infty$, and its lift has $\Bbb Z_p$-coordinates. As wanted.
It remains to check the cases of $p$ among $2,3,5$.
The prime $p=5$. We can work with the equation $Y^2 =X^3 -4320$, since the passage from the given curve to this one involves invertible factors $3,4\in \Bbb Z_p^\times$. The group map $\Bbb F_5^\times\to \Bbb F_5^\times$, $x\to x^3$ is injective. So for each $\bar Y\in\Bbb F_5^\times$ there is an $\bar X\in\Bbb F_5$ with $\bar Y^2=\bar X^3$. Take one such non-singular point, for instance $(1,1)$. Then it can be lifted to a point in $\Bbb Z_5$. Explicitly, lift $X=1$ to $16\in\Bbb Z_5$. We then search for $Y\in\Bbb Z_p$ with $Y^2=16^3-4320=-224=2^4\cdot(-14)$, so we need $-14$ modulo $5$. Well: $$ \small (-14)^{1/2} =(1-15)^{1/2} = 1 -\binom{1/2}1\cdot 15 +\binom{1/2}2\cdot 15^2 -\binom{1/2}3\cdot 15^3 +\binom{1/2}4\cdot 15^4 -\binom{1/2}5\cdot 15^5 +\dots $$ and it is clear that this series converges in $5$-adic numbers.
The prime $p=3$. We must consider the initial equation, $3y^2=4x^3-10$. It turns out that we have a solution for $x=1$. We need $y^2=-2$, and $$ \small (-2)^{1/2} =(1-3)^{1/2} = 1 -\binom{1/2}1\cdot 3 +\binom{1/2}2\cdot 3^2 -\binom{1/2}3\cdot 3^3 +\binom{1/2}4\cdot 3^4 -\binom{1/2}5\cdot 3^5 +\dots $$ and the series converges in $\Bbb Z_3$.
The prime $p=2$. We must consider the initial equation, $3y^2=4x^3-10$. It has no solution in $\Bbb Z/4=\Bbb Z_p/4$, so it has no solution in $\Bbb Z_p$. Indeed, assume there is a solution, then $y^2=-3y^2=-(4x^3-10)= 10=2$. But $2$ is not a square modulo four.