Commutative Exponentiation (Posted on 2017-07-07)

	Submitted by Brian Smith
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Assume x is a multiple of y in the equation x^y=y^x. Then let y=v*x and substitute. After a bit of algebra a parametric solution emerges: x=v^(1/(v-1)) and y=v^(v/(v-1)) v=2 or v=1/2 generates the classic pair of 2 and 4. Assume v is rational. Then 1/(v-1) = p/q for some coprime integers p and q. Then x=((p+q)/p)^(p/q) and y=((p+q)/p)^((p+q)/q). If q is not equal to 1 then irrational roots pop back in, so make q=1. Then x=((p+1)/p)^p and y=((p+1)/p)^(p+1) form a nontrivial rational solution to x^y=y^x for any positive integer. This page at cut-the-knot.org discusses this problem.

	Subject	Author	Date
	re(2): One unusual solution	hellenjos	2018-10-31 03:07:22
	re: One unusual solution	warren zephaniah	2017-08-18 05:07:48
	One unusual solution	Steve Herman	2017-07-07 19:15:29
	rest of the solution	Paul	2017-07-07 15:09:38
	rational solutions	Paul	2017-07-07 14:53:40
	exploration	Charlie	2017-07-07 14:22:36
	Real solutions	Jer	2017-07-07 12:20:19