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Problem

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

a!b! = a! + b! + c²

Solution

Without loss of generality let us assume that b a.

If b = a, then, (a!)² = 2(a!) + c².

(a!)² 2(a!) + 1 = c² + 1
    (a! 1)² = c² + 1

However, as there exists no square which is one more than another square we deduce that b a.

In addition, b 1, otherwise, a! = a! + 1 + c² c² = -1.

Dividing a!b! = a! + b! + c² by b! we get, a! = a!/b! + 1 + c²/b!.

Clearly LHS is integer, and as a!/b! is integer, c²/b! must also be integer. Furthermore, RHS 3, which means that a! 3 a 3.

From a!b! = a! + b! + c², we get a!b! a! b! + 1 = c² + 1.

(a! 1)(b! 1) = c² + 1.

Let us assume that a prime, p 3 mod 4 divides the RHS.

If c² + 1 0 mod p, it follows that c² -1 mod p, and it is clear that p does not divide c.

(c²)^(p1)/2 = c^p1 (-1)^(p1)/2 mod p.

But we are given that p 3 mod 4, so p1 2 mod 4, which means that (p1)/2 will be odd, and (-1)^(p1)/2 = -1.

Hence c^p1 -1 mod p, which is a contradiction, because HCF(c, p) = 1 and by Fermat's Little Theorem, c^p1 1 mod p.

In other words, there exists no prime, p 3 mod 4, which divides c² + 1.

However, for a 4, a! 0 mod 4, and a! 1 3 mod 4; that is, LHS is divisible by a prime, p 3 mod 4.

Hence 1 b a 3, but we have already established that a 3, so we deduce that a = 3 and b = 2.

From a!b! = a! + b! + c², we get, 12 = 6 + 2 + c² c = 2.

That is, the original equation has a unique solution: a = 3, b = c = 2.

Related problems:

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

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

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