Thank you Sue Liu of Madras College, St
Andrews for your solution.
In solving the equation
(1+
1a
)(1+
1b
)(1+
1c
) = 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+
1a
)(1+
1b
)(1+
1c
)
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 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+
1b
)(1+
1c
) =
43
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 +
1c
=
43
.
45
= (1 +
115
)
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+
1b
)(1 +
1c
) =
32
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.