- #1

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

anyone?

- Thread starter msudidi
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- #1

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

anyone?

- #2

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Not sure if there is a more elegant solution though.

- #3

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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

- #4

HallsofIvy

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"Brute force" **is** a method! Please clarify what you are looking for.

- #5

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Doesn't it work for all positive integers c such that [itex]c = \sqrt{a!b! -a! -b!}[/itex]?

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

anyone?

Spit-balling here, we have [itex]a!b! = a! + b! + c^2[/itex]? Doesn't that imply that [itex]\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!}[/itex]. Don't know where I'm going with that...

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