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What are the possible integer values of $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$$ where $x$, $y$, and $z$ are positive integers?

My suspicion is the the only integer values are $3$ and $5$, the former achievable whenever $(x,y,z)=k(1,1,1)$, and the latter achievable when $(x,y,z)=k(1,2,4)$. This suspicion is only based on playing around with the numbers for a while. I tried multiplying out the factors and using divisibility arguments, but nothing came of it.

This question is motivated by this one, so I'm particular interested if the given sum is ever equal to $4$.

EDIT: My suspicion has turned out to be quite wrong. Please see several examples of other solutions in the comments and answers.

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    $\begingroup$ $2,36,81$ yields $41$. $\endgroup$ – barak manos Jun 26 '14 at 19:40
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    $\begingroup$ this could get complicated if it's anything like math.stackexchange.com/questions/402537 $\endgroup$ – mercio Jun 26 '14 at 19:41
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    $\begingroup$ $3,126,196$ yields $66$... $4,9,162$ yields $41$... $4,72,162$ yields $41$... $12,63,98$ yields $9$... $18,28,147$ yields $9$... $24,126,196$ yields $9$... And the list goes on... $\endgroup$ – barak manos Jun 26 '14 at 19:45
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    $\begingroup$ This is apparently one of those questions where a tiny bit of computer programming goes a long way. Your profile says you''re an undergraduate mathematics student, so let me make a pitch for you to learn some simple programming in some easy language like Python. It takes less than five minutes to write a program to exhaustively search all triples with $x,y,z\le 1,000$, and only a few minutes to run it and produce many examples, including barak manos's $(2,36,81)$. $\endgroup$ – MJD Jun 26 '14 at 19:55
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    $\begingroup$ Although this shows there are more than 3 and 5 as possible solutions, if 4 is not possible, we can not show that via exhaustion (e.g. our program). However if 4 is possible, this may be the best way to do it. $\endgroup$ – Cyllindra Jun 26 '14 at 23:42
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Describing all values of $m$ such that the corresponding solution $(x,y,z)$ exists is an open problem. For the reference (quite old though), see the book of Serpinskii, Remark after solution of problem 155. The book is available here: http://www.isinj.com/aime/250%20Problems%20in%20Elementary%20Number%20Theory%20-%20Sierpinski%20(1970).pdf

However, something is known. For example, the equation $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=m$$ has no solution in positive integers $(x,y,z)$ for $m=4n^2$, where $n∈Z$ and $3$ does not divide $n$. On the other hand, if $m=k^2+5$, $k\in\mathbb{Z}$ then our equation has a solution.

The key idea to construct it is to note that is $(a,b,c)$ is a solution of $$a^3+b^3+c^3=mabc,$$ then one can take $x=a^2b, y=b^2c, z=c^2a,$ to produce solution for the given equation. Now, for $m=k^2+5,$ one can easily take $a=2,b=k^2-k+1$ and $c=k^2+k+1.$

Therefore, your special question for $m=4$ is solved. For the reference, A.V. Bondarenko.

Investigation of a class of Diophantine equations. (Russian. English, Ukrainian summary) Ukraïn. Mat. Zh. 52 (2000), no. 6, 831--836;

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One of methods: $p=\dfrac{x}{y}$, $q=\dfrac{y}{z}$, $r=\dfrac{z}{x}=\dfrac{1}{pq}$.

(one of them must be $\le 1$).

$$ p+q+\dfrac{1}{pq}=n, $$

$$ p^2q+pq^2+1-npq=0, $$

(quadratic equation on $q$): $$ p\cdot q^2 + (p^2-np)q+1=0 $$

$$ q_{1,2}=\dfrac{np-p^2\pm\sqrt{p^4-2np^3+n^2p^2-4p}}{2p}. $$

To be $q_{1,2}$ rational, must be

$$ p^4-2np^3+n^2p^2-4p = s^2, \qquad s\in \mathbb{Q}. $$

If $p=\dfrac{a}{b}$, then $(x,y,z)=(2ab, 2b^2, abn-a^2\pm s)$ (after killing all common factors).

Of course, cyclic shift of $x,y,z$ may be applied here.

This way, we can find a few integer values of $n$ for not so large vlues of $x,y,z$ (see table below):

\begin{array}{|c|ll|} \hline n & (x,y,z) & \\ \hline 3 & (1,1,1) & \\ \hline 5 & (1,2,4) & \\ \hline 6 & (4, 3, 18) & = (1\cdot 2^2, 3\cdot 1^2, 2\cdot 3^2) \\ & (9, 2, 12) & = (1\cdot 3^2, 2\cdot 1^2, 3\cdot 2^2)\\ \hline 9 & (12, 63, 98) & =(3\cdot 2^2, 7\cdot 3^2, 2\cdot 7^2) \\ & (18, 28, 147) &= (2\cdot 3^2, 7\cdot 2^2, 3\cdot 7^2) \\ \hline 10 & (175, 882, 1620) & = (7\cdot 5^2, 18 \cdot 7^2, 5\cdot 18^2 ) \\ & (245, 450, 2268) & = (5\cdot 7^2, 18 \cdot 5^2, 7\cdot 18^2) \\ \hline 13 & (1053, 6422, 12996) & =(13\cdot 9^2, 38\cdot 13^2, 9\cdot 38^2)\\ & (1521, 3078, 18772) & = (9\cdot 13^2, 38\cdot 9^2, 13\cdot38^2) \\ \hline 14 & (98, 52, 1183) & = (2\cdot 7^2, 13\cdot 2^2, 7\cdot 13^2)\\ & (338, 28, 637) & = (2\cdot 13^2, 7\cdot 2^2, 13\cdot 7^2)\\ \hline 17 & (1620, 925, 24642) & = (5\cdot 18^2, 37\cdot 5^2, 18\cdot 37^2) \\ & (6845, 450, 11988) & = (5\cdot 37^2, 18\cdot 5^2, 37\cdot 18^2) \\ \hline 18 & (22932, 16055, 379050) \\ & (117325, 7098, 167580) \\ \hline 19 & (25, 9, 405) \\ & (81, 5, 225) \\ \hline 21 & (338, 84, 5733) \\ & (882, 52, 3549) \\ \hline 26 & (12996, 7371, 314678) \\ & (74529, 3078, 131404) \\ \hline 29 & (31347, 336518, 894348) \\ & (49923, 132678, 1424332) \\ \hline 30 & (882, 124, 20181) \\ & (1922, 84, 13671) \\ \hline 38 & (739900, 14341829, 27694870) \\ & (1596070, 3082100, 59741791) \\ \hline 41 & (2, 36, 81) \\ & (4, 9, 162) \\ & (196, 5, 350) \\ & (25, 14, 980) \\ & (3698, 124, 41323) \\ & (1922, 172, 57319) \\ \hline 51 & (1053, 13013, 53361) \\ & (1521, 6237, 77077) \\ \hline 53 & (28, 1323, 1458) \\ & (98, 108, 5103) \\ \hline 54 & (3698, 228, 139707) \\ & (6498, 172, 105393) \\ \hline 57 & (1825900, 32851, 2567110) \\ & (157339, 111910, 8745100) \\ 66 & (3, 126, 196) \\ & (9, 14, 588) \\ \hline 69 & (6498, 292, 303753) \\ & (10658, 228, 237177) \\ & (167580, 4720075, 11488218) \\ & (379050, 922572, 25985255) \\ \hline 83 & (225, 4941, 18605) \\ & (405, 1525, 33489) \\ \hline 86 & (10658, 364, 604513) \\ & (16562, 292, 484939) \\ \hline 94 & (12229083, 132678, 22292452) \\ & (894348, 490617, 82433078) \\ \hline 105 & (24642, 364, 919191) \\ & (16562, 444, 1121211) \\ \hline 106 & (1225, 54, 102060) \\ & (35, 66150, 2916) \\ \hline ... & ... \end{array}


It iv very interesting that all founded values have form

$$ (x,y,z)_1 = (a^2b, b^2c, c^2a);\\ (x,y,z)_2 = (b^2a, a^2c, c^2b). $$

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  • $\begingroup$ Is this true that all solutions of this equation have the following form $$(x,y,z)_1 = (a^2b, b^2c, c^2a);\\ (x,y,z)_2 = (b^2a, a^2c, c^2b).$$? $\endgroup$ – user110661 Sep 2 '14 at 12:49

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