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.
On
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.
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.