The irreducibility of $x^{16}-2x^8+8x+1$ over the rationals 
Question: Is $f(x)=x^{16}-2x^8+8x+1$ irreducible over the rationals?

My attempt: Consider $f(x-1)$ which has every term (except for the highest and constant) divisible by $2$. To apply Eisenstein's criterion, I need to show that the constant term in $f(x-1)$ is divisible by 2 and not divisible by 4.
But the constant term in $f(x-1)$ is $(-1)^{16}-2(-1^8)+8(-1)+1=-8$, which is where I got stuck.
Any Suggestions? Thanks.
 A: To me the simplest way for this one is Perron's criterion used on the reciprocal $f^*(x)=x^{16}+8x^{15}-2x^8+1$. Since $8>1+2+1$, the conclusion follows promptly.
A: Let $f=x^{16}-2x^8+8x+1$.

To show that $f$ is irreducible in $\mathbb{Q}[x]$, it suffices to show that $f$ is irreducible in $\mathbb{Z}[x]$.

Initiating a proof by contradiction, assume $f=gh$, where $g,h\in\mathbb{Z}[x]$ are monic and non-constant.

Since the constant term of $f$ is equal to $1$, the constant terms of $g,h$ are either both equal to $1$ or both equal to $-1$, hence for each of $g,h$, the product of the roots has absolute value equal to $1$.

It follows that each of $g,h$ has at least one root with absolute value at most $1$, hence $f$ has at least two roots with absolute value at most $1$.

But in fact, as we proceed to show, $f$ has only one root with absolute value at most $1$.

Let $D=\{x\in\mathbb{C}{\,:\,}|x|\le 1\}$.

Let $u=8x$ and $v=f-u=x^{16}-2x^8+1$.

Then on $\partial D$ we have
$$
|v(x)|
=
|x^{16}-2x^8+1|
\le
|x|^{16}+2|x|^8+1
=
4 < 8
=
8|x|
=
|u(x)|
$$
hence by Rouche's theorem, $u$ and $u+v$ have the same number of zeros in the interior of $D$.

Thus, since $u$ has only one zero in the interior of $D$, and since $u+v=f$, it follows that $f$ has only one zero in the interior of $D$.

Also, for $x\in\partial D$, we have $|v(x)| < |u(x)|$, so
$$
|f(x)|
=
|u(x)+v(x)|
\ge
|u(x)|-|v(x)| > 0
$$
hence $f$ has no zeros on $\partial D$.

Thus, $f$ has only one root with absolute value at most $1$, which yields the desired contradiction.
