 ## Factorial Equation

#### Problem

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

a!b! = a! + b! + c!

#### Solution

Without loss of generality let us assume that a b and divide through by b!: a! = a!/b! + 1 + c!/b!. As we have integers throughout, c b.

As RHS 3, a! 3 a 3.

Clearly a = b = c would give a! = 3, which has no solutions. Therefore at least one of a, c must exceed b.

But if a b and c b then b+1 will divide a! and c! but not 1, so both a and c cannot exceed b.

If a b and c = b we get a! = a!/c! + 2, and then c+1 would divide a! but not 2.

So we conclude that a = b and c b, giving a! = 2 + c!/a!.

If c a+3, then 3 divides a! and c!/a! but not 2, so c cannot exceed a by more than 2.

Writing a! = 2 + c!/a! as a!(a! 2) = c!, and noting that a c a+3, we get a! c! (a + 3)!. Hence a! 2 is equal to (a + 1)(a + 2) or (a + 1).

If a! 2 = (a + 1)(a + 2), we get a! = a2 + 3a + 4. As LHS is divisible by a, RHS will only divide by a if a = 4, but this does not lead to a solution.

If a! 2 = a + 1, we get a! = a + 3. As LHS is divisible by a, RHS will only divide by a if a = 3 b = 3 and c = 4.

That is, 3!3! = 3! + 3! + 4! is the only solution.

Related problems:

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

Factorial And Power Of 2: a!b! = a! + b! + 2c

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

Problem ID: 216 (09 Mar 2005)     Difficulty: 4 Star

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