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This is my hypothesis and I am not sure if this is a valid one. I need a proof for this. My point is that infinite number of minimas are needed if this is valid, but I am lack of knowledge to prove it.

Any one has a proof ?

Hypotehis:

I want to show that a non-constant polynomial $P(x)$ where $x \in \mathbb{N}$ can not have infinite number of the same constant values in its range.

i.e. for some infinite set of distinct $x_i$ values, where $i \in \mathbb{N}$ and $c$ is a constant, $P(x_i) \ne c $

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2 Answers 2

First, recall that any non-constant polynomial $P(x)$ must have finitely many roots. (That is, finitely many $a \in \mathbb{R}$ such that $P(a) = 0$.)

Now suppose $P(x)$ takes on the value $c$ infinitely many times.

Consider $Q(x) := P(x) - c$. Then $Q(x)$ is a non-constant polynomial with infinitely many roots. Contradiction.

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thank you very nice proof. any one has a proof based on analysis ? –  cingoz recai Nov 1 '12 at 8:44

Here's a sketch of a proof "based on analysis" as the OP requested, although I'm not sure what the point of such a proof is. Let $S$ be the set of roots of $Q(x)=P(x)-c$, and assume that $S$ is infinite.

  1. From the fact that $P(x)\rightarrow\pm\infty$ as $x\rightarrow\pm\infty$, deduce that there must exist a closed interval that contains $S$.
  2. Since this closed interval is compact, $S$ must have a limit point $p$.
  3. Write $Q(x)$ as a "power series" around $p$, and use this to conclude that $Q=0$.
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