# 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.

• Incidentally, $f$ is irreducible mod 19, but I doubt it's the intended way to prove that it's irreducible. I might as well have tried any factorization algorithm in $\Bbb Z[x]$. Aug 12 at 10:16
• could you please collaburate how irreducibility Mod 19 follows!
– Paul
Aug 12 at 10:46
• I just ran an algorithm with the computer until it worked (which it didn't have to), that's why I don't think it's the right way. I mentioned it as a curiousity. Aug 12 at 12:12

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.

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.