If $a\leq b$ then $bc+1$ does not divide $ab$.

Question

Here is my problem:

Let $a,b,c$ be positive integers such that $a\leq b$. Show that $bc+1$ does not divide $ab$.

First I thought I could show that $\frac{ab}{bc+1}<1$, but this is not necessarily. I am trying to express it as $\frac{ab}{bc+1}=x+\frac{p}{q}$ where $x$ is an integer and then show that $\frac{p}{q}\neq 0$, but I am unable to do so.

Answer 1

We have $$\gcd(bc+1,b)=1$$ so if $bc+1$ divides $ab$ then it divides $a$. Thus, there is $d\geq 1$ such that $$a=d(bc+1)\geq bc+1>b.$$

(To prove that $\gcd(bc+1,b)=1$ note that if $p$ divides $bc+1$ and $b$, then it divides $bc$ and hence $p\mid(bc+1)-bc=1$.)

Answer 2

If $bc+1$ divided $ab$, we'd have $\gcd(ab,bc+1) = bc+1$. But using properties of the gcd, \begin{align*} \gcd(ab,bc+1) \quad\,\mid\;\quad&\gcd(a,bc+1) \gcd(b,bc+1) \\ =\quad &\gcd(a,bc+1) \gcd(b,1) \\ =\quad &\gcd(a,bc+1) \\ \le\quad &a \\ \le\quad &b \\ \le\quad &bc \\ <\quad &bc+1. \end{align*} So this cannot occur.