Thank you Sue Liu of Madras College, St Andrews for your solution. In solving the equation

(1 + \frac{1}{a}) (1 + \frac{1}{b}) (1 + \frac{1}{c}) = 2

to find the positive integral solutions we can, without loss of generality, take $a ≦ b ≦ c$ . The required solutions are:

a b c

2 4 15

2 5 9

2 6 7

3 3 8

3 4 5

We can argue using inequalities to find all the possible cases. Let

$E = (1 + \frac{1}{a}) (1 + \frac{1}{b}) (1 + \frac{1}{c})$

Notice that if $a$ is too small then $E$ will be too big and if a is too big then $E$ will be too small so, knowing that a is a whole number, we only have a few possibilities. We use similar reasoning to find $b$ and then calculate $c$ in each case.

If $a = 1$ then $E > 2$ so we know $a \neq 1$ . If $a$ is 4 or more then $E ≦ (5 / 4) (5 / 4) (5 / 4) < 2$ so we know that $a$ is equal to $2$ or $3$ . By similar arguments we can now find all possible solutions.

In the first case, $a = 2$ and we have to find positive integers $b$ and $c$ to satisfy

$F = (1 + \frac{1}{b}) (1 + \frac{1}{c}) (1 + \frac{1}{c}) = \frac{4}{3}$

If $b$ is equal to 1, 2 or 3 then $F > 4 / 3$ so we know b is at least 4. If b is 7 or more then $F ≦ (8 / 7) (8 / 7) = (64 / 49) = (4 / 3) (48 / 49) < 4 / 3$ so we know that b must be 4, 5, or 6.

Taking b = 4 we have

$1 + \frac{1}{c} = \frac{4}{3} . \frac{4}{5} = (1 + \frac{1}{15})$

So $c = 15$ . Similarly if $b = 5$ then $c = 9$ and if $b = 6$ then $c = 7$ .

In the second case $a = 3$ and we have to find positive integers $b$ and $c$ to satisfy

$G = (1 + \frac{1}{b}) (1 + \frac{1}{c}) = \frac{3}{2}$

Again we find all the possible cases from considering inequalities. If $b$ is equal to 1 or 2 then $G > 3 / 2$ and if $b$ is 5 or more then $G ≦ (6 / 5) (6 / 5) = (3 / 2) (24 / 25) < 3 / 2$ so we know that $b = 3 or 4$ and we can calculate the corresponding values $c = 8$ and $c = 5$ .

It is not hard work here to 'exhaust' all the possible cases and we have now found all the solutions.