How do I prove the following for the given congruence

Question

If I have prime($x$) s.t, $x\equiv 5\mod 8$,

How can I prove that $$xa^2 + 6b^2 \equiv 1 \mod 3x $$ has no solutions? I know this will involve Legendre somehow but since $8$ is not an odd prime I don't know how to proceed further

Answer 1

Here is my approach. I will use $p$ instead of $x$ to express as a prime. Since $p \equiv 5 \mod 8$, $p \neq 3$. Let assume that the equation is solvalbe for $a, b$. From the equation, one has a systems of congruence: \begin{equation} pa^2 + 6b^2 \equiv 1 \mod p \tag 1 \\ pa^2 + 6b^2 \equiv 1 \mod 3 \end{equation} For the second equation in $(1)$, we infer $pa^2 \equiv 1 \mod 3$ because $3| \space 6$. Moreover, having $a^2 \equiv \{0, 1\}\mod 3$, we obtain $p \equiv 1\mod 3$, with the fact that $p$ is odd, we conclude $p \equiv 1 \mod 6$.

Now, for the first equation, which is equivalent to $6b^2 \equiv 1 \equiv 5p + 1 \mod p$. Having $5p + 1 \equiv 0 \mod 6$, yeids: $$b^2 \equiv \frac{5p+1}{6} \mod p\Rightarrow \left(\frac{\frac{5p+1}{6}}{p}\right) = 1$$ By Euler's criterion: $$ \left(\frac{5p+1}{6}\right)^{\frac{p-1}{2}} \equiv 1 \mod p \Rightarrow 6^{\frac{p-1}{2}}\equiv (5p+1)^{\frac{p-1}{2}}\equiv 1 \mod p $$ Therefore: $$\left(\frac{6}{p}\right) = \left(\frac{2}{p}\right) \left(\frac{3}{p}\right) = 1 \tag 2$$ However, since $p \equiv 1 \mod 3$, by Gauss reciprocity law and addition property: \begin{equation} \left(\frac{3}{p}\right) \left(\frac{p}{3}\right) = \left(\frac{3}{p}\right) = (-1)^{\frac{(p-1)(3-1)}{4}} = 1 \\ \left(\frac{2}{p}\right) = (-1)^{\frac{p^2-1}{8}} = (-1) \end{equation} The last equality holds because $p \equiv 5 \mod 8$. This contradicts $(2)$. $\square$