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?
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1Are $x,y,z$ positive? – Elaqqad Apr 18 '15 at 23:13
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This might help -- it's a little bit simpler than your question but might give some guidance: http://math.stackexchange.com/questions/403036/natural-number-solutions-to-fracxyxy-n-equivalent-to-frac-1x-frac-1y – Brenton Apr 18 '15 at 23:13
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$x,y,z\in \mathbb{Z}$ – Josimar Apr 18 '15 at 23:15
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Would $(1,10,-2)$ be a valid solution, then? – Arthur Apr 18 '15 at 23:22
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Yeah, this is ONE solution. I'm trying to found all integer solutions. Do you have any idea? – Josimar Apr 18 '15 at 23:25
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Some autors has called these equations “Egyptian”. You have actually a cubic equation of three variables so it is not so easy. – Piquito Apr 18 '15 at 23:26
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Hi Elaqqad, if you are not wrong (what I believe without proof) this could be useful really – Piquito Apr 18 '15 at 23:29
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I'm sorry I made a mistake $$(x(3z-5)-5z)(y(3z-5)-5z)=(5z)^2$$ – Elaqqad Apr 18 '15 at 23:34
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The problem with this factorization is that $z$ is a variable so we don't have informations about its prime factors – Josimar Apr 18 '15 at 23:54
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The solution there. http://math.stackexchange.com/questions/450280/erdös-straus-conjecture/831870#831870 – individ Apr 19 '15 at 04:29
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$(4,4,10)$ is an obvious solution. – Lucian Apr 19 '15 at 06:00
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All solutions for $|x|,~|y|\le10^3$ can be found here. – Lucian Apr 19 '15 at 06:30
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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.
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