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)
a) By inspection, abc - 1 > (a-1)*(b-1)*(c-1)
b) But abc - 1 is a multiple of (a-1)*(b-1)*(c-1).
Therefore, abc - 1 >= 2*(a-1)(b-1)(c-1),
since abc - 1 <> (a-1)*(b-1)*(c-1)
Adding 1 to each side and dividing by abc gives
1 >= 2*((a-1)/a)*((b-1)/b)*((c-1)/c) + 1/abc
But if a >= 4, then b >= 5 and c>= 6
And (a-1)/a increases as a increases.
At a minimum, the expression on the right =
2*(3/4)*(4/5)*(5/6) + 1/abc = 1 + 1/abc, which is > 1
So, a cannot be >= 4.
c) Therefore, a can only be 2 or 3
Edited on May 19, 2007, 8:31 pm
Some initial analytics (spoiler)
