(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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# Solving an equation involving factorials

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