Let $n=3^{100}+2$ and assume that $X^2-53$ does not have zeroes in $\mathbb{Z}/ n\mathbb{Z}$. Show that $n$ is not a prime.
I tried solving this problem by assuming that $n$ is a prime (in order to reach a contradiction). Then, using the law of quadratic reciprocity for Legendre symbols, we get $-1=\left(\frac{53}{3^{100}+2}\right)=\left(\frac{3^{100}+2}{53}\right)$. However, I do not know how to proceed from here.
Hint: Start with Fermat's little theorem: $$3^{100} = 3^{52}\cdot 3^{52}\cdot 3^{-4} \equiv 3^{-4}\mod{53}.$$ Now all you have to do is find the inverse of $3$ modulo $53$ (or just note that $3^{-4}$ is clearly a square) and continue with what you've been doing to compute the Jacobi symbol.