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

(1+

)(1+

) = 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+ 1
a
)(1+ 1
b
)(1+ 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 ≠ 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+ 1
b
)(1+ 1
c
)(1+ 1
c
) = 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 + 1
c
= 4
3
. 4
5
= (1 + 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+ 1
b
)(1 + 1
c
) = 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.