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Consider rational functions $F(x)=P(x)/Q(x)$ with $P(x),Q(x) \in \mathbb{Z}[x]$. I'd like to know when I can expect $F(k) \in \mathbb{Z}$ for infinitely many positive integers $k$. Of course this doesn't always happen ($P(x)=1, Q(x)=x, F(x)=1/x$). I am particulary interested in answering this for the rational function $F(x)=\frac{x^{2}+3}{x-1}$.

Vidit Nanda
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A. Pascal
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2 Answers2

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If $F=P/Q$ is integral infinitely often then $F$ is a polynomial.

Write $$P(x)=f(x)Q(x)+R(x)$$ for some polynomial $R$ of degree strictly less than the degree of $Q$. If you have infinitely many integral $x$ so that $P/Q$ is integral then you get infinitely many $x$ so that $NR/Q$ is integral, where $N$ is the product of all denominators of the coefficients in $f$. However $R/Q\to 0$ as $x\to \pm \infty$ so $R\equiv 0$ and so $Q(x)$ is a divisor of $P(x)$.

Now, as pointed out by Mark Sapir below, not all polynomials with rational coefficients take on integer values infinitely often (at integers), but you can check this in all practical cases by seeing if $dF$ has a root $\pmod{d}$, where $d$ is the common denominator of the coefficients in $F$.

Gjergji Zaimi
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    Not quite because $f$ may have rational coefficients. Example: $(2x+1)/3$. –  Sep 07 '10 at 12:51
  • True, I just assumed $f$ had integral coefficients and it is obvious how to reduce to this case, just multiply the equation with the common denominator of the coefficients of $f$! – Gjergji Zaimi Sep 07 '10 at 12:58
  • Still too fast and not quite correct. Example: $(2x+1)/2$. No integer values, but it is a polynomial (over rationals). –  Sep 07 '10 at 13:05
  • Dear Mark, I think that you understood my "only when" as an "if and only if". I understand that not all polynomials in rational coefficients produce infinitely many integers at integers, only the ones that are integer combinations of $\binom{x}{n}$ do! – Gjergji Zaimi Sep 07 '10 at 13:09
  • Actually my last claim is wrong (it refers to polynomials which take integer values at large enough x, oops), let me edit the answer once more. – Gjergji Zaimi Sep 07 '10 at 13:28
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    I've always been bothered by this argument, because it seems a pain to have to use analysis (even in the very weak form that a rational function of negative degree tends to $0$) to prove a purely algebraic fact. (Insert obligatory Fundamental-Theorem-of-Algebra remark here.) Do you know if there is any way to avoid it? – LSpice Sep 07 '10 at 14:18
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    I think the only thing remaining to be said is that if $F$ is a polynomial then it takes on integer values on a finite (possibly empty) union of arithmetic progressions. In particular, it is integral infinitely often or never at all; there's nothing in between. – Gerry Myerson Sep 08 '10 at 00:09
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$(x^2+3)/(x-1)=x+1+(4/(x-1))$ so this question, at least, is easy; you get an integer if and only if 4 is a multiple of $x-1$.

Gerry Myerson
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