I've considered the following equation for positive integers $x,y,z\geq 1$, and for positive integers $n\geq 2$

$$n^{\frac{1}{x}+\frac{1}{y}}+n^{\frac{1}{y}+\frac{1}{z}}=n^{\frac{1}{z}+\frac{1}{x}},\tag{1}$$ where the pattern of exponent is a cyclic combination of those fractions.

Question 1. Is it known from the literature? Alternatively, if isn't in the literature provide discussion for what work can be done to get the solutions of this equation. Many thanks.

If it is a well-known equation refer/comment it and I try to search the solution and read it from the literature.

Computational facts. The solutions $(x,y,z;n)$ that I can get using a Pari/GP program searching in the segment $1\leq x,y,z\leq 40$, and for $2\leq n\leq 40$, are $(x,y,z;n)=(1,2,1;4)$ $(2,6,2;8)$ and $(2,4,2;16)$. If I'm right the requirement/conditions $y-x\mid xy$ and $y-z\mid yz$ arise when I am under the assumption or case to find solutions of the form $\alpha:=\frac{1}{x}=\frac{1}{z}$ from the deduction $$\frac{1}{\alpha-\frac{1}{y}}=\log_2 n.\tag{2}$$

Question 2. In previous question I try to explore the equation $(1)$ that I've considered yesterday, but is it possible to state a more interesting equation than mine? Can you propose a similar and more interesting equation, or system of equations? Many thanks.

I evoke a more interesting equation than mine, I don't know if equations of the previous type are more or less interesting $$n^{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}+n^{\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}}+n^{\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_1}}=n^{\frac{1}{x_4}+\frac{1}{x_1}+\frac{1}{x_2}},$$ with Question 2 I am looking for tips to improve the mathematical content of my Question 1, asking about how to create a more interesing equation similar than $(1)$, you can add special requirements in your proposal.

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    $\begingroup$ The questions reminds me of $a^b=b^a$ where the only nontrivial integer solution is $(a,b)=(2,4) or (4,2)$. However there one can show that b must divide a (or vice versa) and something similar does not work in your case. $\endgroup$ – user100927 Sep 20 '19 at 13:23
  • $\begingroup$ Many thanks @user100927 I hope that this type of diophantine equations will be interesting for more mathematicians in the future. Many thanks again for your remark, and to the user who posted an answer for previous Question 1 and Question 2 $\endgroup$ – user142929 Sep 21 '19 at 14:52

Question 1: Inspired by the ones you found we can see that there are infinitely many solutions as follows: $$(x,y,z;n) = (k-1,\quad k(k-1),\quad k-1;\quad 2^k)$$ for any $k\ge 0$.

Edit re: Question 2: How about instead of $$n^{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}+n^{\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}}+n^{\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_1}}=n^{\frac{1}{x_4}+\frac{1}{x_1}+\frac{1}{x_2}}$$ you make it $$n^{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}+n^{\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}}=n^{\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_1}}+n^{\frac{1}{x_4}+\frac{1}{x_1}+\frac{1}{x_2}},$$ so that it has greater symmetry.

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    $\begingroup$ So in particular there's no Fermat's Last Theorem style limitation on the size of any of the variables. $\endgroup$ – Bjørn Kjos-Hanssen Sep 5 '19 at 7:58
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    $\begingroup$ It is incredible, many thanks for share your statement. $\endgroup$ – user142929 Sep 5 '19 at 10:56
  • $\begingroup$ I was interested today in other type of equations that evoke variants of other equations that are in the literature. My new problem (I don't know if this is in the literature and I add it as a curiosity if you want to study it in your home) is to determine if the equation $$x^{\sqrt{xy}}=y^{\frac{2}{\frac{1}{x}+\frac{1}{y}}},$$ that involves the geometric and the harmonic means in the exponents, has finitely many solutions for integers $x,y\geq 1$. I can to find the first of those as $x=y=1,60,196,509,\ldots$ I hope don't disturb with this comment, I add this message just as curiosity. $\endgroup$ – user142929 Apr 16 at 14:06
  • $\begingroup$ @user142929 that's true whenever $x=y$, right? $\endgroup$ – Bjørn Kjos-Hanssen Apr 16 at 17:35
  • $\begingroup$ Many thanks for your attention, the comment is just for those solutions that I've showed, I don't know if it is easy to prove that the solutions $(x,y)$ have this property $x=y$. Many thanks again for your great answer in this post. $\endgroup$ – user142929 Apr 16 at 18:10

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