acceloq3c

2022-04-23

Showing that ${x}^{n}-a{x}^{n-1}-bx-1$ , for non-negative integers $n\ge 3,a\ge 1$ , b, has no double roots

Aliana Sexton

Beginner2022-04-24Added 20 answers

Let $f\in \mathbb{Z}\left[x\right]$ be given by

$f={x}^{n}-a{x}^{n-1}-bx-1$

where $n\ge 3,a\ge 1,b\ge 0$.

Claim:f has no repeated roots.

Proof:

Suppose otherwise.

Our goal is to derive a contradiction.

First note that if f has a rational root r, then

- $r=-1$

- n is odd.

- $b=a+2$.

- ${f}^{\prime}(-1)=(a+1)(n-2)>0$.

hence no repeated root of f is rational.

Let w be a repeated root of f.

As you showed (but with a minor correction to your result), w must satisfy the equation

$b(n-1){w}^{2}-(ab(n-2)-n)w-a(n-1)=0$

so we can't have $b=0$, else w would be rational.

Hence w is a root of the quadratic polynomial

$p={x}^{2}-\left(\frac{ab(n-2)-n}{b(n-1)}\right)x-\frac{a}{b}$

which must be irreducible in Q[x] (since w is not rational).

Since p is irreducible in Q[x] and w is a common root of p and f, it follows that p is a factor of f in Q[x], hence we can write $f=pq$ for some $g\in Q\left[x\right]$.

But w is a simple root of p (since p is irreducible in Q[x]), and by assumption, w is a repeated root of f, hence from $f=pq$, we get that w is a root of g, so p is a factor of g in Q[x].

It follows that $p}^{2$ is a factor of f in Q[x].

Note that the discriminant of p is positive, so the roots of p are real, and the product of the roots of p is negative, so one of the roots of p is negative, and the other is positive.

But the roots of p are repeated roots of f, contradiction, since by Descartes' rule of signs, f can't have a repeated positive root.

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