- Thread starter
- #1
- Mar 10, 2012
- 835
Let $r(x)\in\mathbb Q(x)$ be a rational function over $\mathbb Q$. Assume $r(n)$ is an integer for infinitely many integers $n$. Then show that $r(x)$ is a polynomial in $\mathbb Q[x]$.
Let $r(x) = \dfrac{p(x)}{q(x)}$, where $p(x),\,q(x)\in \mathbb Q[x]$. Apply Euclid's algorithm in $\mathbb Q[x]$ to see that $p(x) = a(x)q(x) + b(x)$, where $a(x),\,b(x) \in\mathbb Q[x]$ and $\deg(b(x)) < \deg(q(x)).$ Then $$r(x) = a(x) + \frac{b(x)}{q(x)}.$$ Now choose $n$ to be a multiple of the lcm of the denominators of the coefficients of $a(x)$. Then each term in the polynomial $a(n)$ is an integer except perhaps the constant term. Let $c$ be the fractional part of the constant term. If $b(x)$ is not the zero polynomial then by choosing $n$ large enough we can ensure that $|b(n)/q(n)|$ is nonzero, less than $1$ (because $\deg(b(x)) < \deg(q(x))$), and $b(n)/q(n)$ is different from $-c$ and $1-c$. That would mean that $r(n)$ is not an integer.Let $r(x)\in\mathbb Q(x)$ be a rational function over $\mathbb Q$. Assume $r(n)$ is an integer for infinitely many integers $n$. Then show that $r(x)$ is a polynomial in $\mathbb Q[x]$.
That's good. Here's mine.Let $r(x) = \dfrac{p(x)}{q(x)}$, where $p(x),\,q(x)\in \mathbb Q[x]$. Apply Euclid's algorithm in $\mathbb Q[x]$ to see that $p(x) = a(x)q(x) + b(x)$, where $a(x),\,b(x) \in\mathbb Q[x]$ and $\deg(b(x)) < \deg(q(x)).$ Then $$r(x) = a(x) + \frac{b(x)}{q(x)}.$$ Now choose $n$ to be a multiple of the lcm of the denominators of the coefficients of $a(x)$. Then each term in the polynomial $a(n)$ is an integer except perhaps the constant term. Let $c$ be the fractional part of the constant term. If $b(x)$ is not the zero polynomial then by choosing $n$ large enough we can ensure that $|b(n)/q(n)|$ is nonzero, less than $1$ (because $\deg(b(x)) < \deg(q(x))$), and $b(n)/q(n)$ is different from $-c$ and $1-c$. That would mean that $r(n)$ is not an integer.
The conclusion is that $b(x)$ must be $0$ and therefore $r(x) = a(x) \in \mathbb{Q}[x]$.