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I wish to prove that the binary relation $R_1=\{(x,y):x\in\mathbb{Z}^+,y\in\mathbb{Z}^+,x\neq y$ and $x^y=y^x\}$ is a finite set comprising just the two elements (2,4) and (4,2).

One can easily verify that $2^4=16=4^2$, implying that indeed $(2,4)\in R_1$ and $(4,2)\in R_1$, but proving that the set contains no other elements seems to be a little trickier.

I would suppose the proof should begin: Assume $x^y=y^x$, and one would arrive at the conclusion that either $x$ must be $2$ and $y$ must be $4$ or vice versa, but I am not entirely certain how one should proceed from such an assumption.

Your input is highly appreciated.

The problem was found in the book "Rings, Fields and Groups: An Introduction to Abstract Algebra" by R.B.J.T. Allenby (Second edition pp. 58).

Xam
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Heinrich Wagner
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  • my thought is you can also look at x is odd versus x is even cases and see implications on y if you know enough. –  Aug 29 '17 at 23:43
  • This question is giving an eerie sense of déjà vu. I think I've seen another user post a very similar question about these ordered pairs, presumably out of the same book $\ddot\smile$ Unfortunately, I believe it has since been deleted. – gen-ℤ ready to perish Aug 30 '17 at 00:42

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