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If we have a polynomial function, such that one of its coefficients are not an integer, is it possible that for all natural numbers (0 exclusive) it will return a natural number? Please provide example or disproof.

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Yes. Examples include $\frac{x(x+1)}{2}$.

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  • $\begingroup$ Nice. Even $\frac{1}{n} \prod_{0 \le k \le n - 1} (x + k)$ works, which shows there are such polynomials for all degrees $\ge 2$ $\endgroup$ – vonbrand Feb 12 '13 at 2:42
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Since the tag says "number theory," the following is an interesting class of examples: $$P(x)=\frac{x^p-x}{p},$$ where $p$ is any prime. The fact that this is integer-valued for every integer $x$ is Fermat's (little) Theorem.

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If exactly one coefficient is not an integer, then of course what you are asking for is impossible. The integer-valued polynomials of an integer variable are precisely the integer linear combinations of the binomial coefficients, $$\sum a_j{x\choose j}$$ I don't know what conditions on the coefficients will guarantee natural outputs on natural inputs. Certainly it's necessary (but not sufficient) that the leading coefficient be positive.

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