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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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    I couldn't solve it even for particular case $q=4$ :(2012-02-26
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    By "solution", of course, you mean finding $x,y,z \in \mathbb{N}$. Right?2012-02-27
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    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.2012-02-27
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    @Michael, I don't understand: given, say, $q=4$, how does your calculation find $x,y,z$ (or $a,b,c,d$) that work?2012-02-27
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    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.2012-02-27
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    Sorry for delay in answer @J.D , Yes .2012-02-27

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