Showing that xn−axn−1−bx−1, for non-negative integers n≥3,a≥1, b, has no double roots

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Aliana Sexton · Accepted Answer

Let&amp;nbsp;f∈Z[x]&amp;nbsp;be given byf=xn−axn−1−bx−1where&amp;nbsp;n≥3,a≥1,b≥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-&amp;nbsp;r=−1- n is odd.-&amp;nbsp;b=a+2.-&amp;nbsp;f′(−1)=(a+1)(n−2)&amp;gt;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 equationb(n−1)w2−(ab(n−2)−n)w−a(n−1)=0so we can&#039;t have&amp;nbsp;b=0, else w would be rational.Hence w is a root of the quadratic polynomialp=x2−(ab(n−2)−nb(n−1))x−abwhich 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&amp;nbsp;f=pq&amp;nbsp;for some&amp;nbsp;g∈Q[x].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&amp;nbsp;f=pq, we get that w is a root of g, so p is a factor of g in Q[x].It follows that&amp;nbsp;p2&amp;nbsp;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&#039; rule of signs, f can&#039;t have a repeated positive root.

Showing that x^{n}-ax^{n-1}-bx-1, for non-negative integers n \geq 3, a

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