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Does there exists any other solutions of the following Diophantine equation $$zx^2 +xy^2 +yz^2 =xyzt .$$

I found that $$(x,y,z,t) =(s,s,s,3) ,(x,y,z,t)=(s,2s,4s ,5)$$ where $s\in\mathbb{N}$ are solutions os this equation in positive integers.

I would like to ask, does there exists any other solutions of this equation in positive integers?

Is that true that all solutions of this equation are of the form $(x,y,z,t) =(a^2 b ,b^2 c ,c^2 a ,t)$ for some $a,b,c\in\mathbb{N}$?

2 Answers2

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$$ zx^2+xy^2+yz^2=xyzt\implies t=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}. $$ So your problem is equivalent to finding positive integers $x,y,z$ so that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ is also a positive integer. According to the discussion here, that is an open problem. One trivial observation is that the AM-GM inequality says $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq3$ so $t=3$ (as you've found) is as small as $t$ can get.

Kim Jong Un
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The question has also been posted to MO, where I posted this in reply:

The values of $t$ for which a solution exists are tabulated at the Online Encyclopedia of Integer Sequences. There are also references there and links to further information. In particular, there is a reference to the paper, Andrew Bremner and Richard K. Guy, Two more representation problems, Proc. Edinburgh Math. Soc. (2) 40 (1997), no. 1, 1–17, MR1437807 (98d:11037). The review by Michel Olivier says solutions to $n=x/y+y/z+z/x\,(*)$ correspond to rational points on the elliptic curves $Y^2=X^3+(nX+4)^2$; the paper tabulates $n$ for which $(*)$ has a solution.

The OEIS entry also links to this page of Dave Rusin. That in turn links to these two discussions, which take up where Bremner & Guy left off.

Gerry Myerson
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