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Is it true that if a polynomial $f(x)$ takes integer values for every integer $x$ then its coefficients are integers?

I believe that it is. I just need a hint as to how I can prove or disprove this and just in case you were wondering this is not a homework problem.

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    $\begingroup$ Hint Consider $\displaystyle\,{x\choose n} = \dfrac{x(x-1)\cdots (x-n+1)}{n!}\ $ $\endgroup$ – Bill Dubuque Dec 2 '16 at 4:30
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It's false. Take for example $x(x-1)/2$

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  • $\begingroup$ What would be a sufficient condition for a polynomial to have integer coefficeints? $\endgroup$ – The Cryptic Cat Dec 2 '16 at 4:27
  • $\begingroup$ You could also use the example y = $x(x + 1)(x + 2) / 3$. Notice that if $x$ is divisible by 3, then y will be an integer. If x is not divisible by 3, then either $x + 1$ or $x + 2$ are. $\endgroup$ – Kaynex Dec 2 '16 at 4:30

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