What are the possible solutions for the equation a!b! = a! + b! + c! ?

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Homework Help Overview

The discussion revolves around the equation a!b! = a! + b! + c!, with participants exploring possible solutions in positive integers. The subject area includes factorial properties and combinatorial reasoning.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss specific factorial equations and properties, questioning the equality of different factorial expressions. There is an exploration of the implications of rearranging the original equation and attempts to find integer solutions.

Discussion Status

Some participants have provided insights into specific cases and examples, while others are seeking further clarification on the generality of their findings. There is an ongoing exploration of potential solutions without a clear consensus on the completeness of the results.

Contextual Notes

Participants mention the need to calculate the radius of convergence and reference specific factorial properties, indicating that certain assumptions or definitions may be under consideration.

Mattofix
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Homework Statement



Just a quick on - this is a little bit out of a bigger question.

Does ( 2(n+1))! = 2!(n+1)! ) ?

Do you know of anywhere on the net where i can find factorial properties?

Thanks
 
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Mattofix said:

Homework Statement



Does ( 2(n+1))! = 2!(n+1)! ) ?

Do you know of anywhere on the net where i can find factorial properties?

Thanks

No they are not equal.
If you write out those factorials explicitly you should see why
See http://en.wikipedia.org/wiki/Factorials" if you want more info
 
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Did you TRY anything at all? If n= 1, (2(n+1))!= (2(2))!= 4!= 24 while 2!(n+1)! = 2(2)= 4. You can't get simpler than that.
 
yeah - i did after i posted - too hasty - sorry
 
so I am meant to calcultae the radius of convergence and i have got to

... lZl(2(n+1))!/(2n)!(n+1)^2what shall i do with the (2(n+1))!/(2n)! ?
 
(2n)!= (2n)(2n-1)... (n+1)! so 2(n+1)!/(2n)!= 2/[(2n)(2n-1)...(n+2)!]. what happens to that as n goes to infinity?
 


Hi, similar kind of question.

Find all solutions in positive integers a; b; c to the equation
a!b! = a! + b! + c!

I have rearranged and got (a!-1)(b!-1) = c!+1

And the only solutions I can find are a=3 b=3 c=4 but I can't be sure that they are the only ones. How would I go about finding other solutions?
 

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