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I started with the idea of looking for a $b \in \mathbb Z$ where $f(x+b)$ is a polinomyal that I can use Einsestein for $p = 5$. According with $\dbinom{5}k$ are multiples of $5$ I just need the independent term dividing $5$ and not $5^2$. Lets write this term $i(b) = b^{20} +5b^{15}+25b^{10}+125b^5+625$ according to Fermats Theorem $b^5 \equiv b \ (mod 5)$ then $i(b) \equiv b^4 \ mod(5)$ but I can't find the b with this and I do not know how to continue, some advide? Thanks

Jyrki Lahtonen
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    To not have to do anything with degree $20$ polynomials, maybe you can first notice that $f$ is of the form $P(X^5)$ and study $P$ instead? I don't know yet if it'll lead you closer to a solution but at least the degree will be lesser, so might as well. – Bruno B Oct 17 '22 at 07:31
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    Hint - what happens when, instead of setting $y=x+b$ and trying Eisenstein, you try multiplying rather than adding. The coefficients as powers of $5$ are surely suggestive? And you might want to deal with the fact that all the powers of $x$ are divisible by $5$ at the same time. – Mark Bennet Oct 17 '22 at 07:36
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    @BrunoB Rewriting it in the form $P(x^5)$ and analysing $P$ might work, but there is no guarantee. It doesn't work with $x^5-1$, for instance. – Arthur Oct 17 '22 at 07:41
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    @JeanMarie I really don't think it is. For one thing, $\phi_{25}(5x^5)$ has degree 100, and leading coefficient $5^{20}$. – Arthur Oct 17 '22 at 07:45
  • This is the $25$ th cyclotomic polynomial and those polynomials are irreducible over $\mathbb Q$ – Peter Oct 17 '22 at 13:38
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    I meant that the reciporocal polynomial is $\Phi_5(5x^5)$. Besides, it's hard to recognize the initial question because you have changed it a lot. It's never good to do that. One must leave the initila question as it is and add oe, two... edits. – Jean Marie Oct 17 '22 at 15:59
  • Rolling back to the original version. Don't change a question like that. Particularly after three answers have been posted and more than an hour had passed. Earlier you could have a case for copying it incorrectly, but that won't wash this time. – Jyrki Lahtonen Oct 21 '22 at 06:17

3 Answers3

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Because $2$ is a primitive root modulo $25$, the cyclotomic polynomial $\Phi_{25}(x)$ is irreducible modulo two (or in the ring $\Bbb{Z}_2[x]$). Your polynomial reduces to $\Phi_{25}(x)$ modulo two, so it is irreducible in $\Bbb{Z}[x]$. Gauss's Lemma and friends then imply that it is irreducible in $\Bbb{Q}[x]$ as well.

Jyrki Lahtonen
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To note, since there is a lot of material now in the comments.

If you set $5y=x^5$ you obtain $625(y^4+y^3+y^2+y+1)$ and you can proceed from there.

(Jyrki Lahtonen's approach is more general)

Mark Bennet
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    @Arthur $y-1$ has root $y=1$. Also your reduction would take $x^2-25$ to $5(y-5)$ giving $y=5$ and the two roots $x=\pm 5$. A linear polynomial, though it is irreducible, has a root. There is a step to show that any irreducible factor of the original polynomial which is not comprised entirely of powers of $x^5$ could only correspond to a sub-factor of the polynomial in $y$ [eg as would be the case with $x+1$ as a factor of $x^5+1$ if $x^5+1$ were a factor of the original equation] – Mark Bennet Oct 17 '22 at 09:19
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I'll present a self-contained solution without cyclotomic polynomial essentially following Jykri Lahtonen's answer. By Gauss's lemma, it suffices to prove this is irreducible over $\mathbb Z$ hence over $\mathbb F_2$.

Over $\mathbb F_2$, we have the polynomial is just $x^{20}+x^{15}+x^{10}+x^5+1=0$ which is a geometric progression. Note that this implies $\frac{1-\alpha^{25}}{1-\alpha^5}=0$ for any root $\alpha$, hence $\alpha^{25}=1$. Therefore the order of $\alpha$ in the cyclic group $(\mathbb F_2(\alpha))^{\times}$ can only be $1$ (impossible, as $1$ is not a solution to the polynomial), $5$ (impossible, as if $\alpha^5=1$, then $\alpha$ cannot satisfy the polynomial) and $25$ (the only possibility).

Therefore we have $25 \mid 2^n-1$ where $n:=[\mathbb F_2(\alpha):\mathbb F_2]$. Now we can easily check that none of $n<20$ satisfies this relation, hence $n=20$.

Just a user
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