Solving quadratic equation in $Z_{97}$

Question

I need help solving equations in finite field. Specifically i need to solve $3x^2 + 52 = c $ where $c = 11^{32n + 8}$ in $\mathbb Z_{97}$. I am assuming it has to use the fact that $c = 11^{32n + 8}\equiv 1\mod 96 $ which i proved in first part of the problem. How would i approach solving this problem?

Answer 1

Solve for $x^2$ by first multiplying through by $65$:

$$x^2 = 65c+15$$

Case I: If $n=3k$ then by fermat $c=11^{32n+8} = (11^{96})^k\cdot 11^8 = 36$

Then we have $x^2 = 27 \implies x\in \{30, 67\}$

Case II: If $n=3k+1$ then $c=11^{32n+8} = (11^{96})^k\cdot 11^{40} = 62$

Then we have $x^2 = 68 \implies x\in \emptyset$

Case III: If $n=3k+2$ then $c=11^{32n+8} = (11^{96})^k\cdot 11^{72} = 96$ (or $-1$)

Then we have $x^2 = 47\implies x\in \{12, 85\}$