Integer values of $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$?

Solution 1:

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;

Solution 2:

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

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