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

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Discussion Overview

The discussion revolves around solving the equation a!b! = a! + b! + c^2 for positive integers a, b, and c. Participants explore various methods and approaches to find solutions, including brute force and theoretical reasoning.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests a brute force solution with a=2, b=3, c=2, but questions the existence of a more elegant method.
  • Another participant expresses a desire to find a method to relate the multiplication of two factorials to their sums, implying that if this relationship could be established, the value of c would not pose a problem.
  • A different participant proposes that the equation might work for all positive integers c such that c = √(a!b! - a! - b!), indicating a potential generalization.
  • One participant attempts to manipulate the equation algebraically, suggesting a relationship involving products and factorials but expresses uncertainty about the direction of their reasoning.

Areas of Agreement / Disagreement

Participants generally agree on the brute force solution found but express differing views on the methods to derive a more systematic approach. The discussion remains unresolved regarding a definitive method for solving the equation.

Contextual Notes

Participants have not reached consensus on the best approach to relate the factorial multiplication and sums, and there are unresolved mathematical steps in the proposed manipulations.

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 [itex]c = \sqrt{a!b! -a! -b!}[/itex]? :biggrin:

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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