Finding primes p such that $3x^2=2$ has no solution modulo p

Question

I am not sure how to do this. I know about Legendre symbols and reciprocity but how do I deal with the 3 coefficient?

Answer 1

Hint

By the calculation with the Legendre symbol, it has a solution if and only

$$\biggl(\dfrac2p\biggr)=\biggl(\dfrac3p\biggr)$$ By the second complementary law, $$\biggl(\dfrac2p\biggr)=\begin{cases}\phantom{-} 1 & \text{if }p\equiv1,7\mod8,\\ -1& \text{if }p\equiv3,5\mod8\end{cases}. $$
Next, the law of quadratic reciprocity asserts that $$\biggl(\dfrac3p\biggr)=\begin{cases}\phantom{-}\biggl(\dfrac p3\biggr)& \text{if }p\equiv1\mod4,\\[1ex] -\biggl(\dfrac p3\biggr)& \text{if }p\equiv3 \mod 4.\end{cases}$$ Last, the only non-zero square mod. $3$ is $1$.

From these relations, you should be able to find the pairs $(p\bmod 3,p\bmod 8)\in\mathbf Z/3\mathbf Z\times \mathbf Z/8\mathbf Z$ that solve the problem. There will remain to deduce the corresponding values of $p\bmod24$ by the Chinese remainder theorem.