Repeated factor in polynomial co-efficients and the rational root theorem

Question

This is a strange question, but consider the following type of equation, which falls out of a problem I'm working on in a different field:

$(1-n)^3 x^2 + c_{1}nx + c_{2}n = 0$

where $n,c_{1}, c_{2} \geq 0$ and $\in \mathbb{Z}$. By the rational root theorem, I believe that any integer solution of $x$ must be a factor of $c_{2}n$ - but because $n$ is common to every co-efficient bar the leading one, can I reason as follows?

Reasoning: $(1-n)^3x^2$ must be divisible by $n$, which means that unless $n=1$, the co-efficient cannot be divided by $n$ and thus either $n=1$, or $x = wn$.

I'm not sure ifI'm on the right track here - would I instead have to conclude that $x^2 = wn$ or can I make the inference I have above? I have a whole class of polynomials were all terms bar the leading one have a repeated factor, and I'd be grateful for any advice!

Answer 1

if $a$ is the leading coefficient and $n$ the repeated factor, with $\gcd(a,n)$ , call the degree $d,$ we get $$ a x^d + c_1 n x^{d-1} + \cdots + c_{d-1} n x + c_d n \; , \; $$ and we are setting this to $0,$ we get $$ a x^d \equiv 0 \pmod n $$ Next, there is a multiplicative inverse $b,$ that is $ab \equiv 1 \pmod n,$ as we can find a Bezout combination $$ ab - n v = 1. $$ Take $ a x^d \equiv 0 \pmod n $ and multiply both sides by $b,$ so $$ x^d \equiv 0 \pmod n \; \; . \; \; $$

If $n$ is squarefree, you have your conclusion. If $n$ is not divisible by any $m^d$ for $m > 1,$ you still get your conclusion. What if $n = k^d$ for example?

Take $n = 9,$ and

$$ -512x^2 + 9x + 18378 $$ which is $$ -(x-6)(512x+3063) $$

$$ -512x^2 + 9x + 18378 = -(x-6)(512x+3063) $$

Answer 2

The result becomes clearer (to me, at least) if we quantify your reasoning. Given that $$(1-n)^3x^2+c_1nx+c_2n=0,$$ where $c_1$, $c_2$, $n$ and $x$ are integers, and $c_1,c_2,n\geq0$, we see that $$c_2n=(n-1)^3x^2-c_1nx=x\left((n-1)^3x-c_1n\right),$$ so indeed $x$ is a factor of $c_2n$. Similarly $$(1-n)^3x^2=-n(c_1x+c_2),$$ so indeed $n$ is a factor of $(1-n)^3x^2$. As you note $(1-n)^3$ is coprime to $n$, i.e. the two share no common factor greater than $1$, and so $n$ is a factor of $x^2$. That is to say, $x^2=wn$ for some integer $w$.

It does not follow that $n$ is a factor of $x$. For example, for $(c_1,c_2,n)=(13,1,4)$ we get $$0=-27x^2+52x+4=(-27x-2)(x-2),$$ which has $x=2$ as its only integral solution, but $n=4$ is not a factor of $x=2$.