How can we solve a!b! = a! + b! + c^2 for positive integers a, b, and c?

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The equation a!b! = a! + b! + c^2 is being explored for solutions in positive integers a, b, and c. A brute force approach yields a solution of a=2, b=3, c=2, but participants are seeking a more elegant method. There is discussion on how to relate the multiplication of factorials to their sums, suggesting that if a relationship can be established, c would not pose a problem. One participant proposes that c could be expressed as c = √(a!b! - a! - b!). The conversation highlights the challenge of finding a systematic solution to the equation.
msudidi
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Given that a, b, and c are positive integers solve the following equation.

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

anyone?
 
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I found the answer through brute force: a=2, b=3, c=2.

Not sure if there is a more elegant solution though.
 
vorde, thanks for trying, I am getting the same answer too:smile:

but I'm seeking for method to solve it: how to relate multiplication of 2 factorials and their sums? if we can, then c wouldn't be a problem.

there should be a way to solve it:rolleyes:
 
"Brute force" is a method! Please clarify what you are looking for.
 
msudidi said:
Given that a, b, and c are positive integers solve the following equation.

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

anyone?

Doesn't it work for all positive integers c such that c = \sqrt{a!b! -a! -b!}? :biggrin:

Spit-balling here, we have a!b! = a! + b! + c^2? Doesn't that imply that \displaystyle a! = 1 + \frac{a! + c^2}{b!} = 1 + \frac{a(a-1)(a-2)(a-3)...}{b(b-1)(b-2)...} + \frac{c^2}{b!} = 1 + \prod_{k = (b+1)}^a k + \frac{c^2}{b!}. Don't know where I'm going with that...
 

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