# Polynomials with no multiple root

Let $a,d$ be polynomials of $\mathbb Z[X]$ with $\deg a>\deg d\ge0$ and $P$ be a polynomial of $\mathbb Z[X]$. Consider an infinite sequence of integers $(\lambda_n)_n$. Can one assert there exists a $\lambda_n$ such that $$(\lambda_n +1)^2P-(\lambda_n a+d)(a+\lambda_n d)$$ has only simple roots?

EDIT: The polynomials $a$ and $d$ are relatively prime in $\mathbb Q[X]$.

Thanks in advance.

• Are the $\lambda_n$ all different? – Per Alexandersson May 23 '18 at 16:40
• yes, they are all different. – joaopa May 23 '18 at 16:54
• What about $a = x^{3}$, $d = x^{2}$, $P = x^{2}$? – Minseon Shin May 23 '18 at 17:40
• Sorry, I forgot to say that $a$ and $d$ are relatively prime in $\mathbb Q[X]$ – joaopa May 23 '18 at 18:53

## 1 Answer

It can happen that $(\lambda+1)^2P-(\lambda a+d)(a+\lambda d)$ has multiple roots for any integer $\lambda$: namely, if $P=ad$, then \begin{multline*} (\lambda+1)^2P-(\lambda a+d)(a+\lambda d) \\ = (\lambda^2+1)(P-ad) + \lambda(2P-a^2-d^2) = -\lambda (a-d)^2, \end{multline*} so that any root of $a-d$ is a multiple root of $(\lambda+1)^2P-(\lambda a+d)(a+\lambda d)$.

### A little beyond

One can actually classify completely the cases where $\lambda_n$ with the property in question can be found. Namely, from the identity $$(\lambda+1)^2P-(\lambda a+d)(a+\lambda d) = (\lambda+1)^2(P-ad)-\lambda(a-d)^2$$ it follows that if some root of $a-d$ is a multiple root of $P-ad$, then the polynomial in the left-hand side has multiple roots for any integer $\lambda$. On the other hand, if none of the roots of $a-d$ is a multiple root of $P-ad$, then there are only finitely many those $\lambda$ for which the polynomial has multiple roots; this follows from the general fact that if the polynomials $P$ an $Q$ do not have common multiple roots, then there are at most finitely many $\lambda$ for which $P+\lambda Q$ has a multiple root.