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This post (Find all possible values of $x+y+z$.) got me curious. Consider this problem: Let $x,y,z \in \mathbb{N}$ such that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=t$ is an integer.

Which integers $t$ are expressible in this way? (Are there infintely many when $\gcd(x,y,z)=1$?)

Edit: Conjecture: If $b=\gcd(x,y),c=\gcd(y,z),a = \gcd(x,z)$ then $(x,y,z) = (a^2b,b^2c,c^2a)$ and then $xyz=(abc)^3$ is always a cube, and $x/y+y/z+z/x = \frac{a^3+b^3+c^3} {abc}$

Proof: There exist $k,l,m$ such that $x = bak, y=bcl,z=acm$. From this it follows that (by plugging in $x,y,z$ in $x/y+y/z+z/x=t$ and rearranging:

$a \cdot m(tbckl-abk^2-c^2lm) = k \cdot b^2cl^2$ (1)

Since $\gcd(ak,cl)= \gcd(bk,cm) = \gcd(bl,am)=1$ It follows by the equation (1) that $k|a$ and $a|k$ hence $a=k$. Similarily we get: $l=b,m=c$ and the conjecture is proved. Hence: $(x,y,z) = (a^2b,b^2c,c^2a)$

Also, I found the OEIS sequence: http://oeis.org/A072716

  • If gcd(x,y,z) =1 then $x^2z + y^2x + z^2y =axyz$ but $x^2z + y^2x + z^2y$ is not divisible by x, y or z so there are no solutions. Unless $x =y =z=1$ in which case there is one solution if a = 3. – fleablood Jul 28 '17 at 18:13
  • What about $(x,y,z) = (1,2,4)$? Then $a=5$ –  Jul 28 '17 at 18:17
  • Oh, wait. gcd(x,y,z) doesn't mean gcd(x,y) = gcd(y,z) = gcd(x,z) = 1. Which can't happen. but we could have gcd(x,y) =d gcd(x,z)= e and gcd(y,z) =f with d,e,f, mutually relatively prime. Which I think is very limiting. – fleablood Jul 28 '17 at 18:18
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    See here: https://math.stackexchange.com/questions/848718/integer-values-of-fracxy-fracyz-fraczx for closely related question. – Oleg567 Jul 28 '17 at 21:24
  • One of the answers to the question you linked shows that any integer of the form $2m^2-n^2$ is expressible That gives an infinite set – Ross Millikan Jul 29 '17 at 03:04

1 Answers1

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the problem seems very familiar; here are some coprime solutions. There is more freedom for solutions than initially appears. However, the prime factorizations of $x,y,z$ are fairly rigid, only two or three primes are involved for these early solutions, if three primes one of them is $2.$

Well, reached the first on with four primes.

I think there is one type of rigidity that holds for the solutions found so far. I did not program this check, I have looked at the list myself. It seems that, when a prime $p$ divides one of $x,y,z,$ then it always divides a second one (but not all three, I demand gcd one). Further, if the smallest nonzero exponent of $p$ is $t,$ then the other entry has $p^{2t},$ so that the triple has $1, p^t, p^{2t},$ and $xyz$ is divisible by $p^{3t}.$

This suggests a much quicker program, but not exhaustive. Take tw or three or four or five primes. For each one, either don't use it or pick a small minimal exponent $t.$ Assign factors $1, p^t, p^{2t},$ in one of the six possible orders to $x,y,z.$ Do the same for the other primes chosen, $q$ with exponent $u,$ then $r$ with exponent $v.$ See if a solution occurs. 

 a         x     y     z
 3         1     1     1  :::      1 =   1  ;    1 =   1  ;    1 =   1 
 5         4     1     2  :::      4 =  2^2 ;    1 =   1  ;    2 =  2
 6        12     9     2  :::      12 =  2^2 3 ;    9 =  3^2 ;    2 =  2
 6        18     4     3  :::      18 =  2 3^2 ;    4 =  2^2 ;    3 =  3
41        81     2    36  :::      81 =  3^4 ;    2 =  2 ;    36 =  2^2 3^2
 9        98    12    63  :::      98 =  2 7^2 ;    12 =  2^2 3 ;    63 =  3^2 7
 9       147    18    28  :::      147 =  3 7^2 ;    18 =  2 3^2 ;    28 =  2^2 7
41       162     4     9  :::      162 =  2 3^4 ;    4 =  2^2 ;    9 =  3^2
66       196     3   126  :::      196 =  2^2 7^2 ;    3 =  3 ;    126 =  2 3^2 7
19       225    81     5  :::      225 =  3^2 5^2 ;    81 =  3^4 ;    5 =  5
41       350   196     5  :::      350 =  2 5^2 7 ;    196 =  2^2 7^2 ;    5 =  5
19       405    25     9  :::      405 =  3^4 5 ;    25 =  5^2 ;    9 =  3^2
66       588     9    14  :::      588 =  2^2 3 7^2 ;    9 =  3^2 ;    14 =  2 7
14       637   338    28  :::      637 =  7^2 13 ;    338 =  2 13^2 ;    28 =  2^2 7
41       980    25    14  :::      980 =  2^2 5 7^2 ;    25 =  5^2 ;    14 =  2 7
14      1183    98    52  :::      1183 =  7 13^2 ;    98 =  2 7^2 ;    52 =  2^2 13
53      1458    28  1323  :::      1458 =  2 3^6 ;    28 =  2^2 7 ;    1323 =  3^3 7^2
10      1620   175   882  :::      1620 =  2^2 3^4 5 ;    175 =  5^2 7 ;    882 =  2 3^2 7^2
Will Jagy
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  • $\quad 3, 5, 6, 9, 14,19, 41, 66,\ldots$ OEIS doesn't know this sequence ..........Just found $9, 11826, 21316, 2369$ I mean $\quad\dfrac{9}{11826}+\dfrac{11826}{21316}+\dfrac{21316}{9}=2369\quad$ – Raffaele Jul 28 '17 at 20:12
  • @Raffaele right. In the order my program is going, it takes a while to find a new $a$ value, but that is happening. The sequence is now $3,5,6,9,10,14,19,41,66,$ and I suspect more small values will be filled in eventually – Will Jagy Jul 28 '17 at 20:21
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    Also the Big One follows your rule $\quad x=3^2,;y=2; 3^4 ;73,;z=2^2 ;73^2,;a=23; 103$ – Raffaele Jul 28 '17 at 20:45
  • @Raffaele it occurred to me that a different type of search could be conducted by demanding that my "rule" about the exponents hold, and then (this is the messy program part) all possible $x,y,z$ using a fixed set of primes with an assigned smallest nonzero exponent for each, and ordering them in all ways. Not too bad for just two primes and small exponents. – Will Jagy Jul 28 '17 at 20:51