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Here is my question:

Solve the equation $3m^2\equiv 1\pmod{p}$, where $p$ is some prime.

This seems to be closely related to quadratic residues. For example, if we were to solve $x^2\equiv -1\pmod{p}$, we can conclude that the equation has solutions if and only if $p\equiv 1\pmod{4}$, using the famous Euler's criterion, $$\left(\frac{-1}{p}\right)\equiv (-1)^{\frac{p-1}{2}}\pmod{p}.$$

However, since now the square is multiplied by $3$, I had no idea what to do. I tried multiplying the inverse of $3$ modulo $p$ (assume $p\neq 3$), to get $m^2\equiv 3^{-1}\pmod{p}$ but I don't know if I did anything wrong or is this even the right path to go.

Can anyone help? Thanks in advance.

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Multiply through by 3, getting $(3m)^2\equiv{}3\mod{p}$.

Then there are clearly only solutions if 3 is a quadratic residue. Suppose it is, and $3\equiv{a^2}$. Then $3m\equiv{}\pm{a}$.

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