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?
a!b! = a! + b! + c^2
anyone?
How can we solve a!b! = a! + b! + c^2 for positive integers a, b, and c?
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Vorde
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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.
Not sure if there is a more elegant solution though.
msudidi
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vorde, thanks for trying, I am getting the same answer too
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
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
Science Advisor
Homework Helper
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"Brute force" is a method! Please clarify what you are looking for.
Mandelbroth
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msudidi said:
Doesn't it work for all positive integers c such that [itex]c = \sqrt{a!b! -a! -b!}[/itex]?
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...