Problem 1 « Mathsmaster’s Blog

Problem 1

Problem: Find all positive integers $(a,b,c)$ such as

$\frac{ab-1}{c},\frac{bc-1}{a}$ and $\frac{ca-1}{b}$ are all integers,

Solution:

$\begin{cases}\frac{ab-1}{c}=k\\ \frac{bc-1}{a}=l\\ \frac{ca-1}{b}=m\end{cases}$

Multiplying all equations we get $\frac{a^2b^2c^2-a^2bc-ab^2c-abc^2+ab+bc+ca-1}{abc}=klm$
$\frac{ab+bc+ca-1}{abc}=klm+a+b+c-abc$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}$ is an integer.we have $\frac {1}{a} + \frac {1}{b} + \frac {1}{c} - \frac {1}{abc} = k$ where $k$ is an integer
suppose that one of $a,b,c$ is equal to $1$ for example $c$ then,
$\frac {ac - 1}{b}$ is integer then $\frac {a - 1}{b}$ is integer wich implies $b\leq a - 1$ (*)
$\frac {bc - 1}{a}$ is integer then $\frac {b - 1}{a}$ is integer wich implies $a\leq b - 1$ and from $a\leq b - 1 \leq a - 2$ wich is false. therefore $a,b,c \geq 2$ thus
$k = \frac {1}{a} + \frac {1}{b} + \frac {1}{c} - \frac {1}{abc}\leq \frac {3}{2}$ therefore $k = 1$ or $k = 0$
if $k = 0$ then $\frac {1}{a} + \frac {1}{b} + \frac {1}{c} = \frac {1}{abc}$
$\Leftrightarrow ab + bc + ca = 1$ wich is false, thus, $k = 1$
$\frac {1}{a} + \frac {1}{b} + \frac {1}{c} - \frac {1}{abc} = 1$
$\Leftrightarrow \frac {1}{bc}\left(\frac {bc - 1}{a} + b + c\right) = 1$
and we have $a\geq 2$ therefore,
$1 = \frac {1}{bc}\left(\frac {bc - 1}{a} + b + c\right)\leq \frac {1}{bc}\left(\frac {bc - 1}{2} + b + c\right)$
$\Leftrightarrow bc \leq bc + 1 \leq 2(b + c) \Leftrightarrow \frac {1}{2} \leq \frac {1}{b} + \frac {1}{c}$
suppose that $b\geq c$ therefore, $\frac {1}{2} \leq \frac {1}{b} + \frac {1}{c}\leq \frac {2}{c} \Leftrightarrow 2\leq c\leq 4$
if $c = 4$ then, $\frac {1}{a} + \frac {1}{b} - \frac {1}{4ab} = \frac {3}{4}$
$\Leftrightarrow (3b - 4)(3a - 4) = 13$ wich don’t have any integer solution.
the $c = 2$ or $c = 3$
if $c = 2$
$\frac {1}{a} + \frac {1}{b} - \frac {1}{2ab} = \frac {1}{2}$
$\Leftrightarrow (b - 2)(a - 2) = 3$
then $S = \{2,3,5\}$
if $c = 3$
$\frac {1}{a} + \frac {1}{b} - \frac {1}{3ab} = \frac {2}{3}$
$\Leftrightarrow (2b - 3)(2a - 3) = 7$
then $S = \{2,3,5\}$
finally the only solution for the problem is , $S = \{2,3,5\}$ with all permutation possible.