About xy=yxx^y=y^x

Consider the equation xy=yxx^y=y^x and it is required to find ALL solutions (x,y)(x,y) for this equation,

It is obvious that all ordered pairs (x,x)(x,x), i.e., y=x0y=x \ne 0 is a trivial set of solutions. (Here, I am not sure if x=y=0x=y=0 can be considered as a solution!!)

In addition, the set x=α1α1,y=αxx=\alpha^{\frac{1}{\alpha-1}},y=\alpha x, for some real (not sure if complex works) value of α1,0\alpha \ne 1,0, would also fit in as solutions.

Are there any solutions which do not fit into the above two categories? Is it possible to prove otherwise?

Note by Janardhanan Sivaramakrishnan
3 years, 10 months ago

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There are solutions that do not fit. There will be a negative real solution as long as y=ab>1y=\frac{a}{b}>1, and there might be 2 negative real solutions when y=ab<1y=\frac{a}{b}<1, where aa is an even number and bb is an odd number. I haven't actually studied this in detail yet.

From what I got so far, numerically, there would be 2 negative real solutions when y=aby=\frac{a}{b}, kc<y<1k_c<y<1, 0.752688172043<kc<0.7578947368420.752688172043<k_c<0.757894736842

Julian Poon - 3 years, 10 months ago

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I do not even restricting x,yx,y to be real numbers.

Take α=1\alpha = -1, we get x=i=eπ2i,y=i=eπ2ix=i=e^{\frac{\pi}{2}i},y=-i=e^{\frac{-\pi}{2}i} and

thus xy=eπ2i×(i)=eπ2x^y=e^{\frac{\pi}{2}i \times (-i)}=e^{\frac{\pi}{2}}

and also yx=eπ2i×(i)=eπ2y^x=e^{\frac{-\pi}{2}i \times (i)}=e^{\frac{\pi}{2}}

Janardhanan Sivaramakrishnan - 3 years, 10 months ago

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Related: soumava's algorithm

Agnishom Chattopadhyay Staff - 3 years, 7 months ago

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24=42\Large 2^4=4^2

Nihar Mahajan - 3 years, 10 months ago

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This also fits the pattern. Take α=2\alpha = 2, x=2121=2,y=22=4x=2^{\frac{1}{2-1}}=2,y=2*2=4.

Janardhanan Sivaramakrishnan - 3 years, 10 months ago

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