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I'm clueless about this problem:

Let $P,Q \in \mathbb C[X]$ be polynomials with degree $\geq 1$

Suppose that $P$ and $Q$ have the same set of roots.

Suppose also that $P-1$ and $Q-1$ have the same set of roots.

Prove that $P=Q$

I really don't know how to link multiplicities of roots between the various polynomials at stake.

Any hint is appreciated.

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Gabriel Romon
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  • Let $P$ have $n$ distinct roots and $Q$ have $m$ distinct roots. Then the conditions of the problem imply there are $n+m$ points where $P=Q$. However this could still be smaller than either of their degrees. Maybe it would help if we were given that $\deg(P)=\deg(Q)$? – complexist May 24 '14 at 20:38

1 Answers1

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Suppose that $S$ is the set of roots of $P$, and that $R$ is the set of roots of $P-1$. We have

$$ P(X)=a\prod_{s\in S}(X-s)^{n_s},\quad Q(X)=b\prod_{s\in S}(X-s)^{m_s} $$ and $$ P(X)-1=a\prod_{r\in R}(X-r)^{n'_r},\quad Q(X)-1=b\prod_{r\in R}(X-r)^{m'_r} $$ Now, Clearly $S\cap R=\emptyset$, and every $s\in S$ is a root of multiplicity $n_s-1$ of $P'$ and every $r\in R$ is also a root of multiplicity $n'_r-1$ of $P'$. This implies that $$\eqalign{\deg P-1&=\deg P'\geq \sum_{s\in S}(n_s-1)+\sum_{r\in R}(n'_r-1)\cr &\geq \deg P-|S|+\deg P-|R| } $$ or equivalently $$ \deg P< |S|+|R|. $$ Similarly, we have $$ \deg Q< |S|+|R|. $$ But then, if $T=P-Q$, then $T$ has degree smaller than $|S|+|R|$ and has $|S|+|R|$ roots, namely the elements of $R\cup S$. So, $T$ has to be the zero polynomial.$\qquad\square$

Omran Kouba
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  • Note $\ $ This is Problem 7 from the 16th Putnam competition (1956), and the above answer is the standard published solution, e.g. see the compendium by Gleason et. al. If this is the source of the answer then it should be cited. – Bill Dubuque May 24 '14 at 21:12
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    @BillDubuque, Thank you for the bibliographic reference. It is not the source of the answer. Your remark falttters me. – Omran Kouba May 24 '14 at 21:24