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Let $P(x)$ be a polynomial with real coefficients. $deg(P) \ge 2$. Prove that it

is not possible that whenever $P(x)$ is an integer, $x$ is also an integer i.e., there exists

$x_0 \in \mathbb{R}\setminus\mathbb{Z}$ such that $P(x_0)\in\mathbb{Z}$

I have no idea how to proceed!!!

  • 0
    Your question is not well asked. – 2017-01-01
  • 0
    Tell me what's the doubt? – 2017-01-01
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    I'm having computer problems. I can't write the rest of my answer . So I posted it to see if I could write it by editing it. No luck. So I deleted what I had written so far. – 2017-01-02
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    Hint 1: $P(x+1)-P(x)$ is unbounded. Hint 2: The Intermediate Value Theorem. – 2017-01-02

1 Answers 1

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From Gerry Myerson's hints.

Since $deg(P)\geq 2$, $P(x+1)-P(x)$ is a non constant polynomial, in particular unbounded as a function. Then, for large enough $c\in\mathbb{Z}$ we have $|P(c+1)-P(c)|>1$. Hence there is an integer $n$ such that $P(c)

In any case, P is continuous so by the intermediate value theorem there is $x_0\in \mathbb{R}$ with $c