Congruence equation of the form $ax^2 \equiv 1 \pmod{p}$

Question

I'm trying to solve $13j^2 \equiv 1 \pmod{264}$. Now I've seen the form $x^2 \equiv n \pmod{m}$ but the form $ax^2 \equiv n \pmod{m}$ although it looks like a quadratic congruent equation, I don't seem to be able to follow exactly the same steps as $ax^2 +bx+c \equiv 0 \pmod{p}$. Some help would be appreciated (with detail and explanations). Thank you in advance.

Answer 1

Hint because $\gcd(13, 264)=1$ then $13x+264y=1$ has $(61,-3)$ as a solution, thus $13\cdot 61 \equiv 1 \pmod{264}$ and $$13 j^2 \equiv 1 \pmod{264} \iff 13\cdot61 j^2 \equiv61 \pmod{264} \iff\\ j^2 \equiv61 \pmod{264}$$