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I have the polynomial $f(X)=X^{2n}-2X^{n}+1-p$ where $p$ is a prime number and $n\in\mathbb{N}$. I want to check whether it is irreducible or not over $\mathbb{Q}[X]$.

If $2^{2}\nmid1-p$ then $f(X)$ is irreducible by Eisenstein's Criterion. However, I can't make any progress when I consider the polynomial $f(X)=X^{2n}-2X^{n}+4r, r\in\mathbb{Z}$.

Any hints?

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    $\begingroup$ Another observation: there are no linear factors, since if $y \in \mathbb{Q}$ is a solution to the equation then $y^n - 1 \in \mathbb{Q}$ is a solution to $X^2 - p$. $\endgroup$ Feb 21 '14 at 14:20
  • $\begingroup$ @Dane After factoring out the n in $n(...)$, how can you guarantee that the coefficients inside the brackets will be in $\mathbb{Z}$? $\endgroup$ Feb 22 '14 at 19:12
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You are correct, it is always irreducible.

Your polynomial factors as $AB$ where $A=X^n-(1+\sqrt{p})$ and $B=X^n-(1-\sqrt{p})$. It will suffice to show that $A$ (and therefore $B$ also) is irreducible over $K={\mathbb Q}[\sqrt{p}]$.

Thanks to Karpilovsky’s theorem (many thanks to Bill Dubeque for quoting it here), it will suffice to show the following points :

(1) $c=1+\sqrt{p}$ is not a $m$-th power in $K$, for any $m\geq 2$.

(2) $c$ is not of the form $-4z^4$ with $z\in K$.

Proof of (1) : suppose $1+\sqrt{p}=(x+y\sqrt{p})^m$ with $x,y\in{\mathbb Q}$. Then $1-p=d^m$ where $d$ is the rational number $d=x^2-py^2$. So $d$ is a rational root of the monic polynomial $X^m-(1-p)$, so $d$ is an integer. As $1-p<0$, $d$ must be a negative integer and $m$ is odd. Then $p=1-d^m$ is divisible by $1-d>0$, so $1-d$ can only be $1$ (clearly impossible) or $p$. So $1-d=p,d=1-p$ and hence $(1-p)^{m-1}=1$, which occurs only when $p=2$.

We then have $1+\sqrt{2}=(x+y\sqrt{2})^m$, $m$ odd and $x^2-2y^2=-1$. Each real number has a unique $m$-th real root, so $x+y\sqrt{2}=(1+\sqrt{2})^{\frac{1}{m}}$ and hence $x-y\sqrt{2}=\frac{x^2-2y^2}{x+y\sqrt{2}}=(1-\sqrt{2})^{\frac{1}{m}}$. Adding those two last equalities, one obtains

$$ x=\frac{(1+\sqrt{2})^{\frac{1}{m}}+(1-\sqrt{2})^{\frac{1}{m}}}{2} $$

Then, $r=2x$ is both rational and a sum of two algebraic integers, so it must be an integer. Now,

$$ r=(\sqrt{2}-1)^{\frac{1}{m}} \Bigg(\bigg(\frac{\sqrt{2}+1}{\sqrt{2}-1}\bigg)^m-1\Bigg) >0 $$

On the other hand, $\big(\frac{3}{2}\big)^3 > \sqrt{2}+1$ yields $(\sqrt{2}+1)^{\frac{1}{m}} \leq \frac{3}{2}$, and $\sqrt{2}-1 > \big(\frac{1}{2}\big)^3$ yields $1+(\sqrt{2}-1)^{\frac{1}{m}} \geq \frac{3}{2}$. Combining the two, we obtain $r<1$. Finally $r$ is an integer strictly between $0$ and $1$, which is impossible.

Proof of (2) : $1+\sqrt{p}=-4(x+y\sqrt{p})^4$ with $x,y\in{\mathbb Q}$ is clearly impossible as a fourth power cannot be negative.

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  • $\begingroup$ It seems you are assuming $x$ and $y$ to be integers. Am I wrong? $\endgroup$
    – Dune
    Apr 6 '14 at 10:33
  • $\begingroup$ @Dune No, I am not. In the proof of (1), $d$ is an integer because it is both rational and root of the monic polynomial $X^m-(1-p)$. $\endgroup$ Apr 6 '14 at 10:36
  • $\begingroup$ @Dune And in the proof of (2), a fourth power is never negative (even if what’s inside the fourth power is not an integer). $\endgroup$ Apr 6 '14 at 10:40
  • $\begingroup$ Ok you are right. Thank you for clarifying! $\endgroup$
    – Dune
    Apr 6 '14 at 10:42
  • $\begingroup$ @Dune The latest version of my answer now covers the $p=2$ case as well, so that it constitutes a full answer. $\endgroup$ Apr 6 '14 at 12:26
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Well, the polynomial $f(X)$ can be written as $(X^n-1)^2 - p$. If $f(X)$ is reducible then there must exist a X such that : $$ f(X) = (X^n-1)^2 - p = 0 $$ which means that $$(X^n-1)^2 = p$$ It necessary follows that $(X^n-1)^2 ∈ N$ ergo $X^{n}-1$ is a natural number, which implies that $X ∈ N$ ... unfortunately there exists no prime number that is also a square number, otherwise it would have at least 1 more divisor different from itself and 1...so the polynomial $f(X)$ is irreducible for any $X ∈ Q.$

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  • $\begingroup$ Doesn't this only show that p(X) has no linear factors, it doesn't exclude the possibility that p(X) might have factors of higher degree. $\endgroup$ Feb 23 '14 at 18:03
  • $\begingroup$ sorry, I can't get what you're saying ... @SomethingWitty $\endgroup$
    – sirfoga
    Feb 23 '14 at 18:56
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    $\begingroup$ You claim that if $f(X)$ is reducible then there must exist $Y\in\mathbb{Q}$ such that $f(Y)=0$, which is true when n=1. However, for n>1 you also need to check that there are no quadratic factors, cubic factors...etc. For example, $f(X)=(X^{2}+1)^2$ has no $Y\in\mathbb{Q}$ such that $f(Y)=0$ but is still reducible in $\mathbb{Q[X]}$. $\endgroup$ Feb 23 '14 at 19:47

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