Find all positive integer solutions \((x,y,z)\) that satisfy: \[\frac1x +\frac2y+\frac3z=1\]

Find all positive integer solutions \((x,y,z)\) that satisfy: \[\frac1x +\frac2y+\frac3z=1\]

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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? – Peiyush Jain · 4 years ago

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(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. – Jimmy Kariznov · 4 years ago

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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! – Subhrodipto Basu Choudhury · 4 years ago

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\[ \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1. \] – Tim Vermeulen · 4 years ago

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– Sawarnik Kaushal · 4 years ago

Of course there exists, for example, x = y = z = 6.Log in to reply