Divisibility with a and b
Posted: Sat May 06, 2017 10:12 am
Determine all ordered pairs $(a,b)$ of positive integers for which $\dfrac{b^3-1}{ab-1}$ is an integer.
$\frac{b^3-1}{ab-1}=x$
Firstly notice that if $(a,b)$ is a solution, then $(b,a)$ is also a solution. So, we may assume wlog $a \ge b$ .
If $b=1$, $x=0$ . So, we got a solution $(a,1)$ where $a>1$ .
Now consider the case $b \ge 2$ . $x>0$. Notice that $x \equiv 1 (mod b)$ . So, $x=nb+1$ . As $a \ge b$ , $nb+1=\frac{b^3-1}{ab-1} \le \frac{b^3-1}{b^2-1} = \frac{b^2+b+1}{b+1} < b+1$ implies $n<1$ . So, $n$ must be $0$ . $\frac{b^3-1}{ab-1}=1$ implies $a=b^2$
Firstly notice that if $(a,b)$ is a solution, then $(b,a)$ is also a solution. So, we may assume wlog $a \ge b$ .
If $b=1$, $x=0$ . So, we got a solution $(a,1)$ where $a>1$ .
Now consider the case $b \ge 2$ . $x>0$. Notice that $x \equiv 1 (mod b)$ . So, $x=nb+1$ . As $a \ge b$ , $nb+1=\frac{b^3-1}{ab-1} \le \frac{b^3-1}{b^2-1} = \frac{b^2+b+1}{b+1} < b+1$ implies $n<1$ . So, $n$ must be $0$ . $\frac{b^3-1}{ab-1}=1$ implies $a=b^2$
