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Is it true that for every rational $q \geq 3$ , the following equation has a solution over $\mathbb N$ ?

$$q=\frac{x}{y} +\frac{y}{z} + \frac{z}{x}$$

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  • $\begingroup$ I couldn't solve it even for particular case $q=4$ :( $\endgroup$ – Mahan Feb 26 '12 at 16:00
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    $\begingroup$ By "solution", of course, you mean finding $x,y,z \in \mathbb{N}$. Right? $\endgroup$ – user2468 Feb 27 '12 at 0:00
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    $\begingroup$ Note that it's enough to show that there are three rational numbers with product 1 and sum $q$. If we have $x/y = a/b$ and $y/z = c/d$ with $a,b,c,d$ integers, then $z/x = bd/ac$. We can then write $q = (ac)/(bc) + (bc)/(bd) + (bd)/(ac)$ to give a solution of the form you seek. $\endgroup$ – Michael Lugo Feb 27 '12 at 2:07
  • $\begingroup$ @Michael, I don't understand: given, say, $q=4$, how does your calculation find $x,y,z$ (or $a,b,c,d$) that work? $\endgroup$ – Gerry Myerson Feb 27 '12 at 2:33
  • $\begingroup$ It doesn't. But let's say, for example, that I had $q = 731/210$, and I had rational numbers whose product is 1, say $2/3, 5/7, 21/10$, which add up to $731/210$. Then I would be frustrated because the numerators and denominators don't match up. But here we have $a = 2, b = 3, c = 5, d = 7$ and so we can write $10/15 + 15/21 + 21/10$, which has the matching numerators and denominators. $\endgroup$ – Michael Lugo Feb 27 '12 at 3:54
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The problem

$N=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$

with $N,x,y,z \in \mathbb{Z}$ was considered by Andrew Bremner and Richard Guy in "Two more representation problems" published in the Proceedings of the Edinburgh Mathematical Society, vol. 40 pp.1-17 in 1997. An online copy is available here. They showed solutions only occurred for those $N$ where the elliptic curve

$t^2=u^3+N^2u^2+8Nu+16$

has rank at least $1$.

For small $N>0$, the first solution is for $N=6$, with $x=18$, $y=4$ and $z=3$.

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  • $\begingroup$ $N=5$ has a solution with $x=4, y=1, z=2$. +1 for the reference, though. $\endgroup$ – Aryabhata Feb 28 '12 at 18:49
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    $\begingroup$ Sorry!!!! N=5 is an outlier in the sense that its elliptic curve has rank 0 but torsion $\mathbb{Z}/6\mathbb{Z}$, whereas generally the torsion is $\mathbb{Z}/3\mathbb{Z}$. The order 6 torsion points give the quoted solution. $\endgroup$ – Allan MacLeod Feb 28 '12 at 19:45
  • $\begingroup$ Allan, thanks for clarifying! $\endgroup$ – Aryabhata Feb 28 '12 at 19:55
  • $\begingroup$ Thanks for answer , I'll study that paper though It seems beyond of my knowledge . Can someone find next $N$ with desired property ? (I'm not familiar with elliptic curve) $\endgroup$ – Mahan Feb 28 '12 at 22:11
  • $\begingroup$ Wonderful! It is another reason for me to try to learn something about elliptic curves. $\endgroup$ – user940 Mar 1 '12 at 20:05
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Maple code that searches for solutions in specific range :

for q from 4 to 30 do
for x from 1 to 200 do
for y from 1 to 200 do
for z from 1 to 200 do
if x/y+y/z+z/x=q then
print(q,x,y,z);
end if;
end do;
end do;
end do;
end do;

For $~q=9~$ ;$~(x,y,z)=(12,63,98)$

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