# The Elementary number theory problem

I have to prove the following: if $a$, $b$, $n \in \mathbb N$ and $\frac{a^k}{b^{k-1} n^k} \in \mathbb Z$ for every $k \in \mathbb N$ then $\frac {a}{bn} \in \mathbb Z$. I was said that this is an easy exercise in number theory, but i am somewhat stuck on it... Please help.

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Consider $(\forall k\ge1)\,bx^k\in\Bbb Z\implies x\in\Bbb Z$ with $x=\frac{a}{bn}$. – blue May 25 '14 at 19:45
Oh, in this reformulation the problem becomes really obvious! – Elensil May 25 '14 at 20:32

Suppose to the contrary that $\frac{a}{bn}$ is not an integer.
Then there a prime $p$ such that the exponent of $p$ in (the prime power factorization of) $bn$ is greater than the exponent of $p$ in $a$. Let the exponent of $p$ in $a$ be $s$, and the exponent in $bn$ be $t$. Then $t\gt s$.
If $b$ is not divisible by $p$, then the exponent of $p$ in $a^k$ is less than the exponent of $p$ in $b^{k-1}n^k$, so $\frac{a^k}{b^{k-1}n^k}$ is not an integer.
Let the exponent of $p$ in $b$ be $u$. We must have $u\ge 1$.
Note that $\frac{a^k}{b^{k-1}n^k}=b\frac{a^k}{b^kn^k}$. The exponent of $p$ in this is $u +(ks-kt)$. If $s\lt t$, then for any large enough $k$, we have $u+k(s-t)\lt 0$, meaning that $b\frac{a^k}{b^kn^k}$ is not an integer.