So I have to prove that $n = 3^{100} + 2$ is not a prime number while we assume that $X^2 - 53$ has no zeroes in $\mathbb{Z}/n\mathbb{Z}$.
Because we are working with quadratic reciprocity in this chapter, I assumed that $\big(\frac{53}{n}\big) = -1$ and by the law of quadratic reciprocity, we know that $\big(\frac{n}{53}\big) = -1$. However, I have no clue how I could use this to prove that $n$ is not prime.
If you're looking for a proof that the true $(3^{100}+2|53)=+1$, here is my approach:
Multiply be $3^4$, a known square, so:
$(3^{100}+2|53)=(3^{104}+3|53)$
where $2×3^4=162\equiv 3\bmod 53$. The exponent on $3$ in the large term is now a multiple of $52$ forcing $3^{104}\equiv 1$ By Fermat's Little Theorem. Thereby
$(3^{100}+2|53)=(1+3|53)=(2^2|53)=+1$
but you found that a prime number for $(3^{100}+2)$ should have given the Legendre symbol $-1$. As an old hit song says, this is how it is when doves cry.
Even though the above Legendre symbol is $+1$, what causes $X^2-53=0$ to have no solutions in $\mathbb{Z}/n\mathbb{Z}$ is a prime factor $p$ of $n=3^{100}+2$ for which $(53|p)=-1$. The above proof does not identify any such factors, but the factor $37121$ quoted by others has this property.