Determine all possible triplet(s) (a, b, c) of positive integers, with a ≤ b ≤ c, that satisfy this equation.
a*b+b*c+c*a - (a+b+c) = 21
1) A cannot be greater than 2.
Let f(a,b,c) = a*b+b*c+c*a - (a+b+c)
f(3,3,3) = 18, so (3,3,3) is not a solution,
f(3,3,4) = 23, which is already too big. Since the products increase faster than the sum, a must be 1 or 2
2) If a = 1, then solving for c in terms of b gives c = 22/b
The only solutions where a = 1 are (1,1,22) and (1,2,11)
3) If a = 2, then solving for c in terms of b
gives c = (24/(b+1)) - 1.
The only solutions where a = 2 are (2,2,7) and (2,3,5).
(Edited to fix copy-and-paste error. Thanks Dej Mar!)
Edited on June 8, 2010, 9:36 am