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I am looking for a way to factorize $x^{38} - 1$ polynomial in $F_5$ . So far I have: $x^{38} - 1 = (x^{19} - 1)(x^{19} + 1) = (x - 1)(x^{18} + x^{17}... + 1)(x + 1)(x^{18} - x^{17} + ... + 1)$

I can show that $(x^{18} + ... + 1)$ is irreducible due to minimal polynomial of primitive root of 19 in $F_5$ having degree 18(precisely $x^{18}+...+1$).

But I have no idea what to do with $x^{18} - x^{17}... +1$. I am not sure if it is irreducible or not.

Heisenbruh 2
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  • $19\mid 5^9-1$ so the 19th root of unite resides in $\Bbb{F}{5^9}$, and hence it's minimal polynomial over $\Bbb{F}_5$ has degree nine only. But once you can factor the cyclotomic polynomial $\Phi{19}(x)$, the other factor is simply $\Phi_{19}(-x)$, and a factorization follows. You can use the law of quadratic reciprocity to conclude that $5$ is a quadratic residue modulo $19$, so degree $9$ is the worst you can expect. – Jyrki Lahtonen May 23 '23 at 10:26
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    See this old answer of mine for a bit of theory. – Jyrki Lahtonen May 23 '23 at 10:27
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    A "simpler" case factored. I use scare quotes, because there may be a trick that simplifies the process. A calculation similar to that in the linked thread shows that the constant terms of each of the monic degree nine factors of $\Phi_{19}(x)$ are both equal to $-1$. Another trick may say more... – Jyrki Lahtonen May 23 '23 at 10:40

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