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I'm trying to solve the diophantine equation $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{5}$ for all $x,y,z \in \mathbb{Z}$. I did this. Since that the diophantine equation is symmetric, we can suppose that $x\leqslant y \leqslant z$ so $xy \leqslant 5z$ then x+y \leqslant 1$ e $x<0$. But I can no go further. Any suggestions?

Josimar
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1 Answers1

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We may suppose that $0\lt |x|\le |y|\le |z|$. Then, since one has $$\frac 35=\left|\frac 1x+\frac 1y+\frac 1z\right|\le \left|\frac 1x\right|+\left|\frac 1y\right|+\left|\frac 1z\right|\le \frac{3}{|x|},$$ one has $$|x|\le 5\Rightarrow x=-5,-4,-3,-2,-1,1,2,3,4,5.$$ This should make it easier to solve the equation.

mathlove
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