# number-theory-Solve-in-Z-1-x-1-y-1-xy-1-4-

Question Number 142514 by mnjuly1970 last updated on 01/Jun/21
$$\:\:\: \\$$$$\:\:\:\:\:\:\:\:\:\:…..\:{number}\:\:{theory}….. \\$$$$\:\:\:\:\:\:\:{Solve}\:{in}\:\mathbb{Z}\:: \\$$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{xy}}\:=\frac{\mathrm{1}}{\mathrm{4}}\:….? \\$$$$\:\:\:\:\:……… \\$$
Answered by ArielVyny last updated on 01/Jun/21
$${conditions}\:{x}\neq\mathrm{0}\:\:{y}\neq\mathrm{0}\:\:{xy}\neq\mathrm{0} \\$$$$\frac{\mathrm{1}}{{x}}={t}\:\:\:\:\frac{\mathrm{1}}{{y}}={u} \\$$$${t}+{u}+{tu}=\frac{\mathrm{1}}{\mathrm{4}} \\$$$${t}\left(\mathrm{1}+{u}\right)=\frac{\mathrm{1}}{\mathrm{4}}−{u} \\$$$${t}=−\frac{{u}−\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}+{u}} \\$$$$−\frac{{u}−\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}+{u}}+{u}−\left(\frac{{u}−\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}+{u}}\right)×{u}=\frac{\mathrm{1}}{\mathrm{4}} \\$$$$−\frac{\mathrm{4}{u}−\mathrm{1}}{\mathrm{1}+{u}}+{u}−\left(\frac{\mathrm{4}{u}−\mathrm{4}}{\mathrm{1}+{u}}\right)×{u}=\mathrm{1} \\$$$$−\frac{\mathrm{4}{u}−\mathrm{1}}{\mathrm{1}+{u}}+\frac{{u}\left(\mathrm{1}+{u}\right)}{\mathrm{1}+{u}}−\left(\frac{\mathrm{4}{u}^{\mathrm{2}} −\mathrm{4}{u}}{\mathrm{1}+{u}}\right)=\mathrm{1} \\$$$$−\mathrm{4}{u}+\mathrm{1}+{u}\left(\mathrm{1}+{u}\right)−\left(\mathrm{4}{u}^{\mathrm{2}} −\mathrm{4}{u}\right)=\mathrm{1}+{u} \\$$$$−\mathrm{4}{u}+\mathrm{1}+{u}+{u}^{\mathrm{2}} −\mathrm{4}{u}^{\mathrm{2}} +\mathrm{4}{u}=\mathrm{1}+{u} \\$$$$−\mathrm{3}{u}^{\mathrm{2}} =\mathrm{0}\:\:\:\:{u}=\mathrm{0} \\$$$${t}=−\frac{{u}−\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}+{u}} \\$$$${t}=−\frac{−\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}}\:\:\:\:\:\:\:\:{t}=\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{{x}} \\$$$$\\$$$$\\$$$$\\$$
Answered by ArielVyny last updated on 01/Jun/21
$${it}\:{is}\:{possible}\:? \\$$
Answered by ArielVyny last updated on 01/Jun/21
$${i}\:{think}\:{is}\:{not}\:{possible} \\$$
Answered by ajfour last updated on 01/Jun/21
$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{xy}}=\frac{\mathrm{1}}{\mathrm{4}} \\$$$$\Rightarrow\:\mathrm{4}{x}+\mathrm{4}{y}+\mathrm{4}={xy} \\$$$$\left({x}−\mathrm{4}\right)\left({y}−\mathrm{4}\right)=\mathrm{20} \\$$$$=\mathrm{1}×\mathrm{20}=\mathrm{2}×\mathrm{10}=\mathrm{4}×\mathrm{5} \\$$$$\Rightarrow\:\left({x},{y}\right)=\left(\mathrm{5},\mathrm{24}\right)\:,\:\left(\mathrm{6},\mathrm{14}\right)\:,\:\left(\mathrm{8},\mathrm{9}\right) \\$$
Commented by mnjuly1970 last updated on 01/Jun/21
$${thsnks}\:{alot}… \\$$$$\:\:\left(\mathrm{2},−\mathrm{6}\right)\:,\:\left(\mathrm{6},−\mathrm{2}\right),\left(−\mathrm{16},\mathrm{3}\right)\:,\left(\mathrm{3},−\mathrm{16}\right) \\$$$$\:\:\:\:{for}\:{example}: \\$$$$\:\:\:\:\left(\mathrm{2},−\mathrm{6}\right) \\$$$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{6}−\mathrm{2}−\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{4}}\:..\checkmark \\$$$$\\$$
Answered by mnjuly1970 last updated on 01/Jun/21
$$\:\:{solution}: \\$$$$\:\:\:\frac{{y}+{x}+\mathrm{1}}{{xy}}=\frac{\mathrm{1}}{\mathrm{4}} \\$$$$\:\:\:\:\:\:\:\:\mathrm{4}{x}+\mathrm{4}{y}+\mathrm{4}−{xy}=\mathrm{0} \\$$$$\:\:\:\:{x}\left(\mathrm{4}−{y}\right)=−\mathrm{4}{y}−\mathrm{4} \\$$$$\:\:\:\:\:{x}=\frac{−\mathrm{4}{y}−\mathrm{4}}{\mathrm{4}−{y}}=\frac{\mathrm{4}\left({y}+\mathrm{1}\right)}{{y}−\mathrm{4}}=\mathrm{4}.\frac{{y}−\mathrm{4}+\mathrm{5}}{{y}−\mathrm{4}} \\$$$$\:\:=\mathrm{4}+\frac{\mathrm{20}}{{y}−\mathrm{4}}\:\:….{x},{y}\:\in\mathbb{Z} \\$$$$\:\:\:\:\:\:{y}−\mathrm{4}=\pm\mathrm{1}\:\:,\:{y}−\mathrm{4}=\pm\mathrm{2}\:\:,\:{y}−\mathrm{4}=\mathrm{4} \\$$$$\:{y}−\mathrm{4}=\mathrm{5}\:,{y}−\mathrm{4}=\pm\mathrm{10} \\$$$$\:{y}−\mathrm{4}=\pm\mathrm{20} \\$$$$\:\:\:,……{easy}… \\$$$$\\$$