Factorial And Power Of 2
Given that a, b, and c are positive intgers, solve the following equation.
a!b! = a! + b! + 2c
Without loss of generality let us assume that a b and divide through by b!: a! = a!/b! + 1 +2c/b!, giving integers throughout.
As each term on the RHS is integer, RHS 3 a! 3 a 3.
For b 2, b! will contain a factor of 3, and so cannot divide 2c. Thus b=1 or b=2.
If b=1, we get a! = a! + 1 + 2c, which leads to 2c + 1 = 0: no solution.
If b=2, we get a! = a!/2 + 1 + 2c1, which leads to a!/2 = 1 + 2c1.
If a 3, a!/2 is even, so 2c1 = 1. But then we get a!/2 = 2: no solution.
If a = 3, 3 = 1 + 2c1 c = 2.
Hence we obtain the only solution: 3!2! = 3! + 2! + 22.
Factorial Symmetry: a!b! = a! + b!
Factorial Equation: a!b! = a! + b! + c!
Factorial And Square: a!b! = a! + b! + c2