Hi guys,
find all solutions of equation a + b + c = a . b . c, where are positive integers.
Well, by trial and error method I found out that 1+2+3 = 1.2.3
Probably there is no other solution, but how can I proove that? I tried case analysis, but unsuccessfully. Any idea? Thanks!
another nice proof (I'm not an author):
Let a ≤ b ≤ c be positive integers, s.t. equation a+b+c = a.b.c holds.
If a > 1, then 4c ≤ a.b.c = a+b+c. So 3c ≤ a+b, what contradicts a ≤ b ≤ c.
So a= 1. Now we have to solve 1+b+c = b.c. Then b.(c-1) = 1+b+c - b = 1 + c, (b-1).(c-1) = 1 + c - (c-1) = 2.
So (b-1).(c-1) = 2. Hence the only solution satistying assumptions is b = 2, c = 3.