$a+b+c+ab+bc+ac=abc+1$ Eşitliği

[stuart clark]

determine all triplets (a,b,c) in a+b+c+(ab+bc+ca)=abc+1. Where a,b,c are positive integer

[Lokman Gökçe]

WLOG let's assume that 1 < a < b < c.

for a = 1, then given equation 1 + b + c + b + c + bc = bc + 1 and there is no solution.

for a = 2, then given equation 1 + 3b + 3c = bc or (b - 3)(c - 3) = 10. Hence (a, b, c) solutions are (2, 5, 8), (2, 4, 13) and all the permutations.

for a = 3, then given equation 1 + 2b + 2c = bc or (b - 2)(c - 2) = 5. Hence (a, b, c) solutions are (3, 3, 7) and all the permutations.

for a = 4, there is so solution of the equation that satisfy 1 < a < b < c.

Therefore let's assume that 5 < a. Now a + b + c + ab + ac + bc = abc + 1 > 5bc + 1 > 5bc.

-> a + b + c + ab + ac > 4bc

-> a + b + c > 2bc

-> (3/2).(b + c) > 2bc

-> 6c > 4bc

-> b < 2. contradiction!.

So, there is no solution for 5 < a. All solutions are (2, 5, 8), (2, 4, 13), (3, 3, 7) and the permutations.

[stuart clark]

Thanks scarface