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 $F(x)=\frac{x^{2}+3}{x-1}$.
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2$\begingroup$ See mathoverflow.net/questions/30204/… $\endgroup$– David E SpeyerSep 7, 2010 at 12:47
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$\begingroup$ Heh, just finished doing that. $\endgroup$– Cam McLemanSep 7, 2010 at 13:08
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$\begingroup$ In the second line, I think that you should delete $[x]$. $\endgroup$– John BentinSep 7, 2010 at 13:11
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2 Answers
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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 $F$.
answered Sep 7, 2010 at 12:47
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1$\begingroup$ Not quite because $f$ may have rational coefficients. Example: $(2x+1)/3$. $\endgroup$– user6976Sep 7, 2010 at 12:51
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$\begingroup$ 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 $f$! $\endgroup$ Sep 7, 2010 at 12:58
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$\begingroup$ Still too fast and not quite correct. Example: $(2x+1)/2$. No integer values, but it is a polynomial (over rationals). $\endgroup$– user6976Sep 7, 2010 at 13:05
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2$\begingroup$ 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? $\endgroup$– LSpiceSep 7, 2010 at 14:18
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2$\begingroup$ 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. $\endgroup$ Sep 8, 2010 at 0:09
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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 $x-1$.
answered Sep 7, 2010 at 12:34
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1$\begingroup$ It seems odd to leave off the punchline that this happens only for x=2,3, or 5. $\endgroup$ Sep 7, 2010 at 13:11
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$\begingroup$ @Cam, I wanted to leave something for OP to do. @L Spice, OP asked for positive integer arguments, so -1 for 0, -1, and -3. $\endgroup$ Sep 8, 2010 at 0:02
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1$\begingroup$ @Gerry, oops, sorry. I assume that negative points for a negative answer is a net positive somehow? $\endgroup$– LSpiceSep 11, 2010 at 6:35