In a past exam paper This question came up I want to see if I'm doing it correctly .
i) Write $3x^2+6x-1\equiv0 \mod 53$ in the form $y^2\equiv d \mod 53$
ii) Determine whether $d$ is a quadratic residue $\mod 53$
iii) does $5x^2-3x+2\equiv 0\mod53$ have solutions
My attempt :
i) Seems as how $53$ is an odd prime and $(3,53)=1$ then $(4.3,53)=(12,53)=1$
So the solutions to $3x^2+6x-1\equiv0 \mod 53$ are the same as the solutions to $12(3x^2+6x-1)\equiv0 \mod 53$ which we can rearrange as $(6x+6)^2\equiv(36+12) \mod 53$
ii) Here we can use Euler's lemma $48^{(p-1)/2}=1 \mod p$ so it is infact a quadratic residue
iii) Here we search for solutions of $20(5x^2-3x+2)\equiv 0 \mod53$ rearranging yields $(10x-3)^2\equiv(-9+40) \mod53$ , so we check if $31 \mod53$ is a quadratic residue again using Euler's lemma, and we get $33^{26}=-1\mod53$ so it's not a quadratic residue and hence this equation has no solutions $\mod53$ .
Finally if everything here seems okay then was Euler's lemma the most efficient method to check for quadratic reciprocity or would you recommend another one ?
$48=4^2\times3$ so the quadratic character of $48$ modulo $p$ is the same as that of $3$. To find the quadratic character of $3\bmod{53}$, I'd use quadratic reciprocity, rather than Euler's lemma. Since $53\equiv1\bmod4$, the quadratic character of $3\bmod{53}$ is the same as the quadratic character of $53\bmod{3}$, which is the quadratic character of $2\bmod{3}$. But $2$ is a quadratic nonresidue modulo $3$, so $48$ is a quadratic nonresidue modulo $53$.
I don't see where the $33$ comes from in iii), but in any event I think quadratic reciprocity beats Euler here, too (at least for hand calculations).