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Solving an equation involving factorials

  1. Aug 19, 2011 #1
    Splitting the reciprocal of a factorial into a sum of reciprocals of positive integer

    I'm interested in finding all positive integers x, y such that [itex]{1 \over x} + {1 \over y} = {1 \over N!}[/itex], [itex]N \in \mathbb{N}[/itex]. I think it's best to gather as many properties of solutions as possible, to make this problem as computationally simple as possible. I'm not sure if there's a closed form solution for (x, y) here. This is trivial for small N, but let's take N > 2.

    If x, y are solutions for a given N!, we immediately get [itex]{{x + y} \over {xy}} = {1 \over N!} \implies {{xy} \over {x + y}} = N![/itex]. Since the right-side is an integer, we get [itex]x + y \mod xy = 0[/itex]. Also, x = y always gives the solution x = y = 2N!, so let's assume without losing generality that x > y > 1.

    Anything else ...? Maybe even a rapid solution?
    Last edited: Aug 19, 2011
  2. jcsd
  3. Aug 19, 2011 #2
    Without loss of generality, we can assume that x>y., But then

    [tex]\frac{2}{y}\geq \frac{1}{x}+\frac{1}{y}=\frac{1}{N!}[/tex]

    So [tex]2N!\geq y[/tex].

    So this shows that there are a finite number of solutions, and that all solutions must have

    [tex]y\in \{1,2,...,2N!\}[/tex]

    Of course, this is only a small step and won't help in solving the general case.
  4. Aug 19, 2011 #3
    No, the number of solutions is a pretty cool result. Thanks!
  5. Aug 19, 2011 #4
    Re: Splitting the reciprocal of a factorial into a sum of reciprocals of positive int

    For a given N, whenever [itex]{-yN! \over N! - y}[/itex] is an integer you should have a solution. Just solved for x.
  6. Aug 19, 2011 #5
    OK, let's delve some deeper. I suggest reading http://mathcircle.berkeley.edu/BMC4/Handouts/elliptic/node4.html [Broken]

    In general, you want to know the integer solutions to

    [tex]N! x+N! y-xy=0[/tex]

    You already know such a solution (0,0). Now there's an easy way to find all the rational solutions (of which there are infinitely many).

    We consider all lines through (0,0) with rational slope. These lines have the form y=tx or x=0. Each of these lines will intersect our conic in another point, which will be rational. In fact, let's calculate the intersection of y=tx and our conic [tex]N! x+N! y-xy=0[/tex]

    Substiting y=tx in the equation of the conic yields

    [tex] N! x+t N! x- tx^2=0[/tex]


    [tex] x((1+t)N!-tx)=0[/tex]

    thus x=0 or [itex]x=\frac{(1+t)N!}{t}[/itex]. The corresponding y is [itex]y=(1+t)N![/itex].

    So all the rational solutions are of the form


    with t rational.
    Last edited by a moderator: May 5, 2017
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