Why is $(x^{2n+1} - (2n+1)x^{n+1} + (2n+1)x^n - 1)/(x-1)^3$ irreducible?

Question

Consider the following polynomial

$$ f_n(x)=x^{2n+1} - (2n+1)x^{n+1} + (2n+1)x^n - 1 $$

Try the first $n$, I find that the $(x-1)^3$ is its factor:

$$ f_1(x)=(x-1)^3,\\ f_2(x)=(x-1)^3 (x^2+3x+1),\\ f_3(x)=(x-1)^3 (x^4+3x^3+6x^2+3x+1),\\ \cdots $$ and that this is the complete factorization of $f_n(x)$ in $\mathbb{Z}[x]$.

But how can you prove that factoring out $(x-1)^3$ gives the complete factorization over $\mathbb{Z}$? That is, how can you prove that $f_n(x)/(x-1)^3$ is irreducible over $\mathbb{Z}$?

PS: I use Mathematica to test $n\leqslant 1000$, all is correct.

Answer 1

You can show that $\forall i\in\{0,1,2\}:f_n^{(i)}(1)=0$ and that $f_n^{(3)}(1)\neq 0$.

Answer 2

$(x-1)^n$ is a factor of the polynomial $f(x)$ with real coefficients iff $f(1)=f'(1)=f''(1)=\cdots=f^{(n-1)}(1)=0$. Here $$f(1)=1-(2n+1)+(2n+1)-1=0,$$ $$f'(1)=(2n+1)-(n+1)(2n+1)+n(2n+1)=0,$$ $$f''(1)=(2n+1)2n-(n+1)n(2n+1)+n(n-1)(2n+1)=0$$ and $$f'''(1)=(2n+1)2n(2n-1)-(n+1)n(n-1)(2n+1)+n(n-1)(n-2)(2n+1) =(2n+1)n(n+1)>0.$$

Answer 3

Here is a proof in the special case that $2n+1$ is prime (and $n>1$).

Let $g_n(x)=f_n(x)/(x-1)^3$ and let $h_n(x)=g_n(x+1)$. By dividing $f_n(x)$ by $x-1$ three times using long division, we find $$g_n(x)=\binom{2}{2}x^{2n-2}+\binom{3}{2}x^{2n-3}+\dots+\binom{n}{2}x^n+\binom{n+1}{2}x^{n-1}+\binom{n}{2}x^{n-2}+\dots+\binom{3}{2}x+\binom{2}{2}.$$ Now note that $$f_n(x)-(x-1)^{2n+1}=x^{2n+1} - 1 - (x-1)^{2n+1} + (2n+1)(x^n-x^{n+1})$$ is divisible by $2n+1$ since every term of $x^{2n+1}-1-(x-1)^{2n+1}$ is divisible by $2n+1$ (the coefficients have the form $\pm\binom{2n+1}{k}$ for $0<k<2n+1$ which is divisible by $2n+1$ since it is prime). It follows that $h_n(x)-x^{2n-2}$ is also divisible by $2n+1$. That is, every term except the leading term of $h_n(x)$ is divisible by $2n+1$. Moreover, the constant term of $h_n(x)$ is \begin{align*} g_n(1) &= \binom{2}{2}+\binom{3}{2}+\dots+\binom{n}{2}+\binom{n+1}{2}+\binom{n}{2}+\dots+\binom{3}{2}+\binom{2}{2} \\ &=\binom{n+2}{3}+\binom{n+1}{3} \\ &=\frac{n(n+1)(2n+1)}{6}. \end{align*} (Alternatively, $g_n(1)$ can be computed as $f_n'''(1)/6$.) In particular, the constant term of $h_n(x)$ is not divisible by $(2n+1)^2$. So, by Eisenstein's criterion for the prime $2n+1$, $h_n(x)$ is irreducible over $\mathbb{Z}$, and hence so is $g_n(x)=h_n(x-1)$.

(This approach sometimes also works when $2n+1$ is not prime; for instance, $h_4(x)$ turns out to still satisfy Eisenstein's criterion for the prime $3$. However, $h_7(x)$ does not satisfy Eisenstein's criterion for any prime.)