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Let $a$, $b$ be integers and $p$ be a prime number such that:

  1. $p$ is greatest common divisor of $a$, b
  2. $p^2$ divides a.

Prove that the polynomial $x^{n+2}+ax^{n+1}+bx^n+a+b$ cannot be decomposed into the product of two polynomials with integer coefficients and degree greater than $1$.

Thomas Andrews
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Snell
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    Hmm so if it's asked to prove something about the polynomial which is not possible, just show it by contradiction. Assume that it's possible to factorise the polynomial into two linear factors with integer coeffs then deduce something to complete the proof. – Ash_Blanc Apr 15 '23 at 15:50
  • Well bro u needed to show your attempt to this question first of all then only u might get help – Ash_Blanc Apr 15 '23 at 16:01
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    I mean, it looks like this is a simple case of Eisenstein's criterion. But they probably want an actual proof of this instance of that result. https://en.wikipedia.org/wiki/Eisenstein%27s_criterion – Thomas Andrews Apr 15 '23 at 16:04
  • Yeah exactly @ThomasAndrews i too thought the exact same thing but since I've not understood it completely yet so I left the idea – Ash_Blanc Apr 15 '23 at 16:09
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    Also, the "greater than $1$" condition is odd, since there are no factorizations into degrees greater than $0.$ – Thomas Andrews Apr 15 '23 at 17:39

1 Answers1

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Assume that the polynomial can be written as a product $$ \left( x^d + \sum_{i=0}^{d-1} g_i x^i \right) \left( x^e + \sum_{i=0}^{e-1} h_i x^i \right) $$ We have $g_0h_0 = a+b$. Since $p^2$ divides $a$ it can't divide $b$ since the gcd of $a$ and $b$ is $p$. So $a+b$ is not divisible by $p^2$, hence precisely one of $g_0$ and $h_0$ is divisible by $p$. We may assume it is $g_0$; hence $h_0$ is not divisible by $p$. Now $g_0h_1 + g_1h_0 \equiv 0\pmod{p}$ is divisible by $p$, so $g_1$ must be divisible by $p$. Then we have $g_0h_2 + g_1h_1+g_2h_0\equiv 0 \pmod{p}$ so $g_2$ is divisible by $p$. Continuing in this way, we find that when we equate the degree $j$ term to zero modulo $p$, we find $g_j$ to be divisible by $p$. Since all coefficients up to the $n+1$-st degree are zero modulo $p$, we find that all the $g_j$ are divisible by $p$.

We will now repeat the same argument, but starting from the top, that is starting from the expression of the coefficient of $x^{n+1}$, which we know to be $a$, and for which the product expansion gives the value $g_{d-1}+h_{e-1}=a$. This then shows $h_{e-1}$ to be divisible by $p$. The second gives $g_{d-1}h_{e-1}+g_{d-2}+h_{e-2}=b$, which shows $h_{e-2}$ to be divisible by $p$. Then we get $h_{e-3}+g_{d-1}h_{e-2}+g_{d-2}h_{e-1}+g_{d-3}=0$ which shows the $h_{e-3}$ to be divisible by $p$. And so on. So each time, when considering the expansion for the coefficient $x^{n+2-j}$, we conclude that $h_{e-j}$ is divisible by $p$. Hence having arrived at the coefficient $x^{n+2-e}$, which is still zero modulo $p$, we conclude that $h_0$ is divisible by $p$, which contradicts our earlier conclusion that $h_0$ is not divisible by $p$.

Hence our assumption that the polynomial $x^{n+2}+ax^{n+1}+bx^n+a+b$ can be written as a product must be wrong. QED

R.P.
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