I went through a similar but slightly different process when solving it, obtaining three solutions which I've verified (they're also on Simon's list): a, b, c = 2, 29, 2 a, b, c = 3, 13, 13 a, b, c = 5, 11, 11 The values for these are 116, 507, and 605 (for both the left and right sides of course). You basically got it, but the final substitutions were wrong for the last two, where you reversed a vs. (b and c): a, b, c = 13, 3, 3 (doesn't work) a, b, c = 11, 5, 5 (doesn't work) neither of which is correct. Tom Tomas Rokicki writes:
For the prime case I'd do this.
Rewrite
a*b + 7*a*c + 15*c = a*b*c
as
a b (c - 1) == c (7a + 15)
If c==2 then this simplifies to
b = 14 + 30/aj
which (going through the prime divisors of 30, which number 3) gives us the solution (a=2, b=29, c=2).
So other than this solution c>2. The left side must be divisible by c, but c-1 never is divisible by the prime c, so either a=c or b=c. We handle the two cases in turn.
If a=c then we have
b = 7 + 22 / (c - 1)
So for b to be integral c must be 1, 2, 11, or 22. None of these give a new solution.
If b=c, then we have:
c = 8 + 15/a
a can be 3 or 5, and this gives c=13, 11 respectively so this gives the solutions (a=13, b=3, c=3) and (a=11, b=5, c=5).
-tom