Here is a proof I came up with in the exam I just took. But I suspect there may be some issues since I think it seems too simple.

Let $p_n(x)$ denote a complex polynomial of order $n$ (coefficients from $\mathbb C$), and suppose $p_n(x)$ has $s$ distinct roots ($s>n$). And let $$q_s(x)=\prod_{i=1}^{s}(x-x_i)$$ where $x_i$ ($i=1,2,\cdots,s$) is one root for $p_n(x)$. Therefore, $q_s(x)$ must divide $p_n(x)$ (is it sufficient to say this?), i.e., $$q_s(x)\mid p_n(x)$$ which requires $$\deg(q_s(x))\le\deg(p_n(x))$$ or $$s\le n$$ which contradicts the hypothesis. So the original statement is right.

Since this question appeared at the end of my exam, I think it wouldn't be this easy to do. Maybe there are indeed some issues in my proof? Can you point it out, or verify that my proof is just ok? Thanks in advance.

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    $\begingroup$ Do you mean distinct roots? $\endgroup$ – Bill Dubuque Apr 10 '15 at 15:10
  • $\begingroup$ @BillDubuque oh yes.. apologies for my poor english. $\endgroup$ – Vim Apr 10 '15 at 15:11
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    $\begingroup$ You must specify the coefficient ring, e.g. $\,x^2-1\,$ has $4$ roots $\,x\equiv \pm1,\pm3$ over $\,\Bbb Z_8 = $ integers mod $8$. $\endgroup$ – Bill Dubuque Apr 10 '15 at 15:16
  • $\begingroup$ @BillDubuque the coefficients are picked from $\mathbb C$, I'll add it to my post. $\endgroup$ – Vim Apr 10 '15 at 15:18
  • $\begingroup$ Yes, then a simple induction using the Factor Theorem can prove what you claim. Take note where the proof cancels differences of distinct roots $\,x_i - x_j$ in order for the induction to work. That's why the above case fails, because $\,\Bbb Z_8$ has nonzero elements that are not cancellable, i.e. it is not an integral domain, e.g. $\,2\cdot 4 = 0\ \ $ Post a proof as an answer and we can give you feedback on it. $\endgroup$ – Bill Dubuque Apr 10 '15 at 15:21

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