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.

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    $\begingroup$ $3^{100}+2$ has 8 divisors $\endgroup$ – Dr. Mathva Apr 14 at 15:25
  • $\begingroup$ You've written at the beginning "So I have to prove that $n=3^{100}+2$ is a prime number". Nevertheless, in the end, you claim that you want "to prove that $n$ is not prime"... $\endgroup$ – Dr. Mathva Apr 14 at 15:28
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    $\begingroup$ $3^{100}+2 \equiv 12^2 \mod 53$ so $\left( \frac{n}{53}\right) = 1$. $\endgroup$ – Robert Israel Apr 14 at 15:32
  • $\begingroup$ Nice, @Robert Israel, but how do we know it is $12^2$? $\endgroup$ – Fareed AF Apr 14 at 15:38
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    $\begingroup$ @FareedAF $53$ is not very big. You can enumerate $1^2 \mod 53$, $2^2 \mod 53$, ..., $26^2 \mod 53$. $\endgroup$ – Robert Israel Apr 14 at 15:52

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:


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


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.

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    $\begingroup$ Right. Had the OP replied to my comment I was going to note $\large \!\bmod 53\!:\,3^{\large 4} n\equiv 2^{\large 2}\,$ so $\,\large\,n\equiv (2/9)^{\large 2},\,$ essentially the same as above (but doesn't require any knowledge of Legendre symbols). $\endgroup$ – Bill Dubuque Apr 14 at 19:49

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