A family of irreducible polynomials

Question

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!

Answer 1

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

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

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

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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.