Let $p$ be a polynomial so that $p(n)$ are integers for all $n\in \mathbb Z$. Is it possible that it has some irrational coefficients?

Remark: A similar question to ask is if all the coefficients are integers: this is not true ($p(x) = \frac{1}{2}x(x+1)$) is an counterexample).

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    $\begingroup$ "Real polynomials that take on only integer values" By this do you mean that $f(\Bbb R)\subseteq \Bbb Z$? Or do you mean that $f(\Bbb Z)\subseteq \Bbb Z$? If it was the first, considering that polynomials are continuous functions if $f$ happened to take on more than one integer value then it must take on all real values between that as well and would therefore not take on only integer values. The only such polynomials then are the constant polynomials, the only such of which that take on integer values have their sole coefficient being an integer and therefore not irrational. $\endgroup$ – JMoravitz Sep 11 '17 at 2:34
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    $\begingroup$ A polynomial's coefficients are uniquely determined by $f(0), f(1), \cdots, f(n)$, where $n$ is the degree of the polynomial. $\endgroup$ – Kenny Lau Sep 11 '17 at 2:37
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    $\begingroup$ No, all the coefficients will be rational. They are a solution of a system with rational coefficients and with a unique solution. $\endgroup$ – orangeskid Sep 11 '17 at 5:22
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    $\begingroup$ @orangeskid You then should probably add, that $\mathbb{Q}$ is a field, and solving such a system could be made through operations in the field $\endgroup$ – dEmigOd Sep 11 '17 at 6:17
  • $\begingroup$ Consider the polynomial $\pi x - \pi x$. The question should clarify that a simplified polynomial is intended to avoid this kind of triviality. $\endgroup$ – Carl Mummert Oct 23 '17 at 23:32

The question has been answered in the comment: Let $n$ be the degree of the polynomial. Write

$$ p(x) = a_0 + a_1 x + \cdots +a_n x^n.$$

Then by substituting $x=0, 1, \cdots, n$,

$$ \begin{pmatrix} 1 & 0 &\cdots &0 \\ 1 & 1^1 & \cdots & 1^n \\ \vdots & \vdots & &\vdots \\ 1 & n^1 & \cdots & n^n\end{pmatrix} \begin{pmatrix} a_0 \\ a_1 \\ \vdots \\ a_n \end{pmatrix} = \begin{pmatrix} p(0) \\ p(1) \\ \vdots \\ p(n) \end{pmatrix}$$

The square matrix on the left is more or less the Vandermonde matrix, which is invertible. Thus $a_0, \cdots, a_n$ are all rational, since

$$ \begin{pmatrix} a_0 \\ a_1 \\ \vdots \\ a_n \end{pmatrix} = \begin{pmatrix} 1 & 0 &\cdots &0 \\ 1 & 1^1 & \cdots & 1^n \\ \vdots & \vdots & &\vdots \\ 1 & n^1 & \cdots & n^n\end{pmatrix} ^{-1}\begin{pmatrix} p(0) \\ p(1) \\ \vdots \\ p(n) \end{pmatrix},$$

$p(i) \in \mathbb Z$ and the inverse matrix has rational coefficients.


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