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About \(x^y=y^x\)

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

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

In addition, the set \(x=\alpha^{\frac{1}{\alpha-1}},y=\alpha x\), for some real (not sure if complex works) value of \(\alpha \ne 1,0\), 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
9 months, 2 weeks ago

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There are solutions that do not fit. There will be a negative real solution as long as \(y=\frac{a}{b}>1\), and there might be 2 negative real solutions when \(y=\frac{a}{b}<1\), where \(a\) is an even number and \(b\) 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 \(y=\frac{a}{b}\), \(k_c<y<1\), \(0.752688172043<k_c<0.757894736842\) Julian Poon · 9 months, 2 weeks ago

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@Julian Poon I do not even restricting \(x,y\) to be real numbers.

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

thus \(x^y=e^{\frac{\pi}{2}i \times (-i)}=e^{\frac{\pi}{2}}\)

and also \(y^x=e^{\frac{-\pi}{2}i \times (i)}=e^{\frac{\pi}{2}}\) Janardhanan Sivaramakrishnan · 9 months, 2 weeks ago

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Related: soumava's algorithm Agnishom Chattopadhyay · 6 months, 2 weeks ago

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Comment deleted 9 months ago

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@Sharky Kesa That's the same as the second condition, just substitute \(\alpha=\frac{1}{\beta -1}\) Julian Poon · 9 months, 2 weeks ago

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@Julian Poon LOL, was trying to figure out why it seemed similar. Thanks. Sharky Kesa · 9 months, 2 weeks ago

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\(\Large 2^4=4^2\) Nihar Mahajan · 9 months, 2 weeks ago

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@Nihar Mahajan This also fits the pattern. Take \(\alpha = 2\), \(x=2^{\frac{1}{2-1}}=2,y=2*2=4\). Janardhanan Sivaramakrishnan · 9 months, 2 weeks ago

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