# Family of irreducible polynomials

Consider the following family of polynomials

$$P_n(X)=\sum_{i=0}^n(n+1-i)X^i,\,n\ge 1$$

Let’s write down the first few

\begin{align} P_1(X)=&X+2\\ P_2(X)=&X^2+2X+3\\ P_3(X)=&X^3+2X^2+3X+4\\ P_4(X)=&X^4+2X^3+3X^2+4X+5 \end{align}

My claim is that this family is a family of irreducible polynomials over $$\Bbb{Z}[X]$$

I proved it for $$n \le 5$$ by the Eisenstein criterion after the change of variable $$X=t+1$$

For $$n=6$$ after the change of variable the polynomial writes as follows

$$Q_6(t)=P_6(X-1)=t^6+8t^5+28t^4+56t^3+70t^2+56t+28$$

And the Eisenstein criterion (only workable $$p=2$$) doesn’t work anymore.

By reducing $$\pmod 7$$, we prove the claim for $$n=6$$.

I tested with Mathematica and irreducibility is checked for $$n\le 150$$

I noticed

$$P_n(X)=XP_{n-1}(X)+n+1$$

But I am struggling to find a generic proof. I have given up on counterexample. Thanks for your help.

• Irreducible over what? Over $\mathbb{Z}[X]$? Over $\mathbb{C}[X]$, this family is definitely not irreducible. – RMWGNE96 May 8 at 10:45
• Over $\Bbb{Z}[X]$ sorry for not being precise. I will edit – marwalix May 8 at 10:47
• Their reciprocals are derivatives of simpler polynomials, but why would that help? – Jyrki Lahtonen May 8 at 10:47
• Speaking of reciprocals: note that $P_n(X)+X^nP_n(1/X)=(n+2)\sum_{i=0}^nX^i$ – RMWGNE96 May 8 at 11:12
• Here is the question dealing with mentioned prime constant coefficient: Explain proof of irreducibility of $x^{p-1} + 2x^{p-2} \dots (p-1)x + p$ – Sil May 13 at 23:04

This is not a complete answer, but summarizes two known provable cases. For simplicity (and better familiarity with existing posts), let $$f_n(x)=x^{n-1}+2x^{n-2}+\dots+(n-1)x+n$$.

Case 1: $$n+1$$ is a prime. As you have noticed, in many cases we can use Eisenstein criterion for $$f_n(x+1)$$. We can show that this will work when $$n+1$$ is a prime. We have $$f_n(x+1)=\sum_{i=0}^{n-1}(n-i)(x+1)^i=\sum_{i=0}^{n-1}(n-i)\sum_{j=0}^{i}\binom{i}{j}x^j=\sum_{j=0}^{n-1}\sum_{i=j}^{n-1}(n-i)\binom{i}{j}x^j$$ and the coefficient at $$x^j$$ is $$[x^j]f_n(x+1)=\sum_{i=j}^{n-1}(n-i)\binom{i}{j}=\sum_{i=0}^{n-1}(n-i)\binom{i}{j}=\binom{n+1}{j+2}$$

Now having $$p=n+1$$ a prime, we have $$p \mid \binom{n+1}{j+2}$$ for $$j+2, i.e. $$j. Also for $$j=0$$ the coefficient becomes $$\binom{n+1}{2}=\frac{n}{2}(n+1)$$ and it is clearly divisible by $$p$$ but not by $$p^2$$, so the polynomial satisfies conditions of Eisenstein criterion for $$p$$ and is therefore irreducible.

Case 2: $$n$$ is a prime. The irreducibility can be also shown for $$n$$ being a prime as discussed in comments. The idea is that all complex roots of the $$f_n(x)$$ lie outside of the unit circle in a complex plane, which can be shown by looking at roots of $$(x-1)f_n(x)=x^n+x^{n-1}+\dots+x-n$$. Then having $$f_n(x)=p(x)q(x)$$ with $$n$$ prime implies that without loss of generality we can take $$p(x)$$ such that $$|p(0)|=1$$. But $$|p(0)|=\prod |x_k|>1$$ gives a contradiction. More details on this can be found here Explain proof of irreducibility of $x^{p-1} + 2x^{p-2} \dots (p-1)x + p$.

Remaining cases are problematic and I could not find anything about this in literature or prove it. I've tried to get more by looking at the roots but it did not get me much, here is at least an observation I stumbled upon - all complex roots of the polynomial lie in annulus $$1<|x_k|<\sqrt[n]{2n+1}$$. First inequality is already proven above. For the second inequality notice that $$(x-1)^2f_n(x)=x^{n+1}-(n+1)x+n$$, and so for a root $$z$$ we have $$n=|z^{n+1}-(n+1)z|=|z||z^n-(n+1)|.$$ Now being interested only in roots $$|z|>1$$, this gives $$n>|z^n-(n+1)|$$, which geometrically means that $$z^n$$ distance from $$n+1$$ is less than $$n$$, so especially $$|z^n|, from which we have $$|z|<\sqrt[n]{2n+1}$$.

Hopefully someone in the future will have an idea on the remaining cases, but notice that this problem was posted in Putnam and there it was only for $$n$$ being prime, so this seems like a hard problem.

• Thanks !! Very helpful – marwalix May 19 at 10:06
• @marwalix You're welcome. Too bad I cannot prove it for all $n$, but this is typical for most of the polynomial families... – Sil May 19 at 10:07
• Something is wrong for $n=6$, $n+1=7$ is prime. The coefficient of $x^5$ is $8$ and according to your formula it should be $1$. As soon as I am home I will check – marwalix May 19 at 10:17
• @marwalix For $n=6$ we get the polynomial $x^5+2x^4+3x^3+4x^2+5x+6$, and for $x+1$ this is $x^5+7x^4+21x^3+35x^2+35x+21$, so that is correct (remember I have written $f_n(x)=x^{n-1}+2x^{n-2}+\dots+(n-1)x+n$, while your original problem used shifted $n$). – Sil May 19 at 10:20
• You’re right. The shift mislead me. – marwalix May 19 at 10:22