$X^{2n} - 2X^n -7$ irreducible over $\mathbb{Q}$? I read something somewhere that would imply $X^{2n} - 2X^n -7$ is irreducible over $\mathbb{Q}$. Is there an easy way to prove this? 
 A: Here’s one way of doing it, likely not the easiest.
First, the ring of integers in $\Bbb Q(\sqrt2\,)$ is $\Bbb Z[\sqrt2\,]$, which is Euclidean and thus a Principal Ideal Domain. I’ll call this ring $R$.
For the base case $n=1$, where we’re talking about the quadratic polynomial $X^2-2X-7$, the roots are $1\pm2\sqrt2$. We can concentrate on the $+$ case.
The element $1+2\sqrt2$ is indecomposable (prime) in $R$—you can see this because its Norm down to $\Bbb Q$ is $-7$, which is indecomposable in $\Bbb Z$.
Now any root of your polynomial $X^{2n}-2X^n-7$ is an $n$-th root of either $\pi=1+2\sqrt2$ or $\pi=1-2\sqrt2$; these are “primes” in $R$, so (by Eisenstein applied in $R$), $X^n-\pi$ is $R$-irreducible, and this means that $\bigl[\Bbb Q(\sqrt2)[\pi]:\Bbb Q(\sqrt2)\bigr]=n$. This shows that $\bigl[\Bbb Q(\sqrt2)[\pi]:\Bbb Q\bigr]=2n$.
What have we seen? That by adjoining a root of $X^{2n}-2X^n-7$, we get an extension of degree $2n$. This shows that the above polynomial of degree $2n$ is irreducible.
A: We can use the following criterion by Osada (see also Methods to see if a polynomial is irreducible )

Let $f(x)=x^n+a_1x^{n-1}+\dots+a_{n-1}x\pm p$ be a polynomial with integer coefficients, where $p$ is a prime. If $p>1+|a_1|+\dots+|a_{n-1}|$, then $f(x)$ is irreducible over $\mathbb{Z}$.

The conditions are satisfied as $7>1+2$. 
Following is a way to prove the irreducibility without invoking the criterion (yet using the same idea as its proof). The proof is based on location of complex roots and the fact that the constant term is a prime, it has already been used for example in Prove that the polynomial $f(X)=X^{2017}+X^{2016}+...+X^3+X^2+2017$ is irreducible in $\mathbb{Z}[X]$. or Explain proof of irreducibility of $x^{p-1} + 2x^{p-2}+ \dots +(p-1)x + p$.
Let $P(x)=x^{2n} - 2x^n -7$ and consider a root $\alpha$, i.e. $\alpha^{2n}-2\alpha^n-7=0$. For the sake of contradiction, assume $|\alpha| \leq 1$. Then we have
$$
7=|\alpha^{2n}-2\alpha^n|\leq|\alpha|^{2n}+2|\alpha|^n\leq 1+2=3,
$$
(using the triangle inequality), a contradiction. Therefore $|\alpha|>1$, i.e. all its roots lie outside of the unit circle.
Now let's assume we have $P(x)=Q(x)R(x)$ for $Q,R\in \mathbb{Z}[x]$. Since $|Q(0)R(0)|=|P(0)|=7$ is a prime, one of the $|Q(0)|$, $|P(0)|$ must equal to $1$, say $|Q(0)|=1$. But notice that for the roots $\beta_i$ of $Q(x)=\prod(x-\beta_i)$ we have
$$
1=|Q(0)|=\left|\prod (-\beta_i)\right|=\prod \left|\beta_i\right| > 1
$$
(since all roots of $Q(x)$ are also roots of $P(x)$, so $|\beta_i| > 1$), a contradiction. So no such factorization is possible, hence $P(x)$ is irreducible over $\mathbb{Z}$, and by Gauss's Lemma it is irreducible also over $\mathbb{Q}$.
