Determine all triplets of integers (a, b, c) satisfying 1 < a < b < c such that abc - 1 is a multiple of (a - 1)(b -1)(c - 1)
Based on my previous analysis and the one by Steve Herman, since a can be only 2 or 3 and all a, b and must be either odd or even, the following is the solution for odd numbers:
a = 3
As Steve Herman has shown:
1 = x * 2/3 * (b-1)/b * (c-1)/c + 1/(3bc)
Let's try minimum solution for b=7 and c = 9 (it can not be 8 since all must be odd):
1 = x * 2/3 * 6/7 * 8/9 + 1/189 = 0.508 * x + 1/189
x >= 2 and 2/3 * (b-1)/b * (c-1)/c is increasing with b, so
0.508 * x > 1 -> discard.
So b<=7
But also, b>3 and b: odd -> b = 5
1 = x * 2/3 * 4/5 * (c-1)/c + 1/(15c)
Let's try minimum solution for c = 17:
1 = x * 2/3 * 6/7 * 16/17 + 1/255 = 0.502 * x + 1/255
x >= 2 and 2/3 * 6/7 * (c-1)/c is increasing with c, so
0.502 * x > 1 -> discard.
So c <= 17
But also, c > 5 and c: odd -> 7 <= c <=15
The only c that satisfies the equation is 15.
So the only solution for odd numbers is 3, 5, 15.
The assumption that x > 1 is not explicitly defined, but it can be proven, it just is quite lengthy. If it's needed, please notify.
Partial solution
