When does a a rational function have infinitely many integer values for integer inputs? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T06:24:14Z http://mathoverflow.net/feeds/question/37972 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37972/when-does-a-a-rational-function-have-infinitely-many-integer-values-for-integer-i When does a a rational function have infinitely many integer values for integer inputs? A. Pascal 2010-09-07T12:24:59Z 2010-09-07T13:39:57Z <p>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 $F(x)=\frac{x^{2}+3}{x-1}$.</p> http://mathoverflow.net/questions/37972/when-does-a-a-rational-function-have-infinitely-many-integer-values-for-integer-i/37974#37974 Answer by Gerry Myerson for When does a a rational function have infinitely many integer values for integer inputs? Gerry Myerson 2010-09-07T12:34:12Z 2010-09-07T12:34:12Z <p>$(x^2+3)/(x-1)=x+1+(4/(x-1))$ so this question, at least, is easy; you get an integer if and only if 4 is a multiple of $x-1$. </p> http://mathoverflow.net/questions/37972/when-does-a-a-rational-function-have-infinitely-many-integer-values-for-integer-i/37975#37975 Answer by Gjergji Zaimi for When does a a rational function have infinitely many integer values for integer inputs? Gjergji Zaimi 2010-09-07T12:47:33Z 2010-09-07T13:39:57Z <p>If $F=P/Q$ is integral infinitely often then $F$ is a polynomial.</p> <p>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 $P(x)$.</p> <p>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 $F$.</p>