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Problem

Given that a, b, and c are positive intgers, solve the following equation.

a!b! = a! + b! + 2^c

Solution

Without loss of generality let us assume that a b and divide through by b!: a! = a!/b! + 1 +2^c/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 2^c. Thus b=1 or b=2.

If b=1, we get a! = a! + 1 + 2^c, which leads to 2^c + 1 = 0: no solution.

If b=2, we get a! = a!/2 + 1 + 2^c1, which leads to a!/2 = 1 + 2^c1.

If a 3, a!/2 is even, so 2^c1 = 1. But then we get a!/2 = 2: no solution.

If a = 3, 3 = 1 + 2^c1 c = 2.

Hence we obtain the only solution: 3!2! = 3! + 2! + 2².

Related problems:

Factorial Symmetry: a!b! = a! + b!

Factorial Equation: a!b! = a! + b! + c!

Factorial And Square: a!b! = a! + b! + c²