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?

  • $\begingroup$ 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

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.


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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