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Well, all is contained in the title. I cannot think of an example of a real polynomial of degree $\geq 1$ that has all coefficients irrational and which have infinite number of integers as its values.

Somehow, I think that there is no such polynomial but I do not know at the moment how to prove that there exist or that there does not exist such a polynomial.

Does such polynomial exist?

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  • $\begingroup$ More interesting is whether it can take on infinitely many integer values at integer (or rational) values of $x$. $\endgroup$ – André Nicolas Mar 25 '16 at 20:48
  • $\begingroup$ @AndréNicolas Yes, indeed. $\endgroup$ – Farewell Apr 5 '16 at 22:34
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Consider $f(x) = \pi x + \pi$. Then for $x= \frac{n}{\pi} -1$ we have $f(x) = n$, thus we can reach any integer.
Furthermore, for any polynomial $f$ of odd degree, we have $\lim_{x \to \infty} f(x) = \pm \infty$ and $\lim_{x \to -\infty} f(x) = \mp \infty$ depending on the sign of the highest degree, thus we can reach any integer.

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Any polynomial of degree greater than or equal to one satisfies what you want. Polynomial functions in $\mathbb{R}$ are continuous. Odd-degree polynomials go to $+\infty$ as $x \rightarrow +\infty$ and $-\infty$ as $x \rightarrow - \infty$ (or a swap of these, depending on the leading coefficient) and even-degree polynomials go either to $+ \infty$ as $x \rightarrow \infty$ (any of them) or $- \infty$ as $x \rightarrow \infty$. Either way, by the intermediate value theorem, we have that all integers above a certain value $M$, or below, depending on how the polynomial behaves at infinity, are reached as a value of the given polynomial.

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