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Does someone has an idea how to prove that the polynomials \begin{align*} P_a=(a+2)(a+1)X^{a+4}-2(a+4)(a+1)X^{a+3}+(a+4)(a+3)X^{a+2}-2(a+4)X+2(a+1) \end{align*} are all of the form $(X-1)^4Q_a$ with $Q_a$ irreducible (over $\mathbb{Q}$)? Sage tell me it's true until 100 but I can't prove it. In fact it's enough for me that it works for an infinity of such polynomial and even that the biggest degree of all its factor tends to infinity. I tried the classical Eisenstein and reduction modulo $p$ but witout success. Thanks for all your ideas!

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  • $\begingroup$ My impression is you should be able to factor out at least some power of $X-1$ without making the expression any more complicated. $\endgroup$ Sep 1, 2019 at 21:07
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    $\begingroup$ You can always factor out $(x-1)^4$, which you can see by the fact that $\frac{\partial^n}{\partial x^n}P_a(x)$ vanishes for $x=1$ for $n=0,1,2,3$. The result of evaluating $\frac{\partial^4}{\partial x^4}P_a(x)$ at $x=1$ is $2a^4+20a^3+70a^3+100a+48=2(a+1)(a+2)(a+3)(a+4)$, which is only zero at $a=-1,-2,-3,-4$, so if you have $a\geq 0$ for instance it is never zero and $P_a=(x-1)^4Q_a$ where $Q_a(1)\neq 0$. To go further, you'll have to perform this division and get a better handle on $Q_a$. Providing small examples of $Q_a$ in your post could help. $\endgroup$
    – KReiser
    Sep 1, 2019 at 22:20
  • $\begingroup$ @KReiser's comment is quite helpful. The idea is to find the Taylor polynomial of $P_a$ around $X=1$. In the previous comment they already found the constant term of the $Q_a$ using the Taylor polynomial. For a fixed $a$ you can easily find the coefficient of the highest degree term as well. Perhaps you can apply Eisenstein's by pushing this approach further. $\endgroup$
    – user347489
    Sep 1, 2019 at 22:52
  • $\begingroup$ I tried to factor but the result is a polynomial that lost its "empty" property. I don't find for the moment a formula for $Q_a$. $\endgroup$ Sep 2, 2019 at 7:21
  • $\begingroup$ Modulo $p=a$, with $Y=X+1$, I find $P_a=Y^4(Y^p+4Y^{p-3}+3Y^{p-4}+1)$ but $Y^p+4Y^{p-3}+3Y^{p-4}+1$ don't have any reason to be irreducible. $\endgroup$ Sep 2, 2019 at 7:23

1 Answer 1

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$$P_a=(a+2)(a+1)x^{a+4}-2(a+4)(a+1)x^{a+3}+(a+4)(a+3)x^{a+2}-2(a+4)x+2(a+1)$$

Offset $a$ down by one:

$$P'_a=(a+1)ax^{a+3}-2(a+3)ax^{a+2}+(a+3)(a+2)x^{a+1}-2(a+3)x+2a$$

Subst $z = x-1$

$$P'_a=(a+1)a(z+1)^{a+3}-2(a+3)a(z+1)^{a+2}+(a+3)(a+2)(z+1)^{a+1}-2(a+3)(z+1)+2a \\ = \sum_i \left[ (a+1)a\binom{a+3}{i} - 2(a+3)a\binom{a+2}{i} + (a+3)(a+2)\binom{a+1}{i}\right]z^i - 2(a+3)z - 6 \\ = \sum_{i\ge 4} \left[ (a+1)a\binom{a+3}{i} - 2(a+3)a\binom{a+2}{i} + (a+3)(a+2)\binom{a+1}{i}\right]z^i \\ %= \sum_{i\ge 4} \left[ (a+1)a \frac{(a+3)!}{(a+3-i)!i!} - 2(a+3)a \frac{(a+2)!}{(a+2-i)!i!} + (a+3)(a+2)\frac{(a+1)!}{(a+1-i)!i!} \right]z^i \\ %= \sum_{i\ge 4} \frac{(a+3)(a+2)(a+1)a - 2(a+3)(a+2)a(a+3-i) + (a+3)(a+2)(a+3-i)(a+2-i)}{(a+3-i)(a+2-i)} \binom{a+1}{i} z^i \\ %= \sum_{i\ge 4} \frac{(a+3)(a+2)(i-2)(i-3)}{(a+3-i)(a+2-i)} \binom{a+1}{i} z^i \\ = \sum_{i\ge 4} (i-2)(i-3) \binom{a+3}{i} z^i \\ $$

So

$$\frac{P'_a}{z^4} = \sum_{j=0}^{a-1} (j+2)(j+1) \binom{a+3}{j+4} z^{j}$$


In fact it's enough for me that it works for an infinity of such polynomial

When $a+3$ is a prime, Eisenstein's criterion works.

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  • $\begingroup$ Thanks! I hope the calculation don't take you 6 months ;-) It's elegant: I finally made it but with Euclidean division (for finding formula) and recurrence for proving it: it was longer. $\endgroup$ Dec 20, 2019 at 17:57
  • $\begingroup$ Did you also prove it just for $a + 3$ is prime, or was your solution more general? If it's more general, I'd be interested to see a self-answer. $\endgroup$ Dec 20, 2019 at 18:40
  • $\begingroup$ No it was not general: it was a subset of the $a=p$ prime (I don't make your change of variable and get $k(a+3-k)(a+2-k)$ as coefficients, I don't see Eisenstein for this so (after simplify by 2) reduced modulo 2 and get $X\Phi_q^4$ in $\mathbb{F}_2[X]$ and there is an infinity of $\Phi_q$ irreducible here. $\endgroup$ Dec 21, 2019 at 12:29

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