A tedious solution would be to use the fact that \(\min \{x,y,z\} \le 6\) and solve 12 different equations in two variables of the form \[\frac{a}{u} + \frac{b}{v} = \frac{p}{q} \] (which I believe should present no problems). I presume you have an elegant way of reducing the tedium?

Since 0 in denominator of a fraction makes it to infinity the equation does not hold any sense. And also fraction is not an integer. There this equation has no integer solution!!!!! According to my view!

You don't have to let any one of them be \(0\), in fact, they are all positive. Also, several fractions can have a sum that is an integer, for example,

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TopNewestA tedious solution would be to use the fact that \(\min \{x,y,z\} \le 6\) and solve 12 different equations in two variables of the form \[\frac{a}{u} + \frac{b}{v} = \frac{p}{q} \] (which I believe should present no problems). I presume you have an elegant way of reducing the tedium?

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Any complete solution would still have to find all of the following 32 solutions:

(2,5,30), (2,6,18), (2,7,14), (2,8,12), (2,10,10), (2,12,9), (2,16,8), (2,28,7), (3,4,18), (3,6,9), (3,12,6), (3,30,5), (4,3,36), (4,4,12), (4,8,6), (5,4,10), (5,10,5), (5,40,4), (6,3,18), (6,4,9), (6,6,6), (6,24,4), (8,4,8), (8,16,4), (10,5,6), (12,3,12), (12,12,4), (14,4,7), (15,6,5), (20,10,4), (30,3,10), (36,9,4)

So, there probably isn't a less tedious way to do it than the method you described.

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Since 0 in denominator of a fraction makes it to infinity the equation does not hold any sense. And also fraction is not an integer. There this equation has no integer solution!!!!! According to my view!

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You don't have to let any one of them be \(0\), in fact, they are all positive. Also, several fractions can have a sum that is an integer, for example,

\[ \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1. \]

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Of course there exists, for example, x = y = z = 6.

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