Question Number 71294 by Mohamed Amine Bouguezzoul last updated on 13/Oct/19
$${solve}\:{in}\:\mathbb{Z}\:\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{1}}{{p}}\:{with}\:{p}\in\mathbb{P} \\ $$
Commented by Rasheed.Sindhi last updated on 13/Oct/19
$$\frac{{x}+{y}}{{xy}}=\frac{\mathrm{1}}{{p}} \\ $$$$\frac{{xy}}{{x}+{y}}={p} \\ $$$${xy}=\left({x}+{y}\right){p} \\ $$$${The}\:{difference}\:{between}\:{this}\:{question} \\ $$$${and}\:\:{Q}#\mathrm{69939}\:{is}\:{that}\:{there}\:{is}\:{z}\in\mathbb{Z}^{+} \\ $$$${instead}\:{of}\:{p}\in\mathbb{P}.{I}-{e}\:{this}\:{question}\:{is} \\ $$$${special}\:{case}\:{of}\:{Q}#\mathrm{69939}. \\ $$
Answered by mr W last updated on 13/Oct/19
$${x}=\mathrm{1}+{p} \\ $$$${y}=\left(\mathrm{1}+{p}\right){p} \\ $$$$ \\ $$$${examples}: \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{30}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$...... \\ $$
Commented by Mohamed Amine Bouguezzoul last updated on 13/Oct/19
$${nice}\:{work}\:{but}\:{there}\:{exist}\:{more}\:{solutions} \\ $$$${for}\:{example}\:\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$
Commented by mr W last updated on 13/Oct/19
$${that}'{s}\:{true}\:{sir}.\:{please}\:{see}\:{Q}\mathrm{69939} \\ $$$${for}\:{more}. \\ $$
Commented by Mohamed Amine Bouguezzoul last updated on 13/Oct/19
$${god}\:{bless}\:{you}\:{sir} \\ $$