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Question Number 142514 by mnjuly1970 last updated on 01/Jun/21

$$$$$$.....numbertheory.....$$$$Solvein\mathbb{Z}:$$$$\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}=\frac{1}{4}....?$$$$.........$$

Answered by ArielVyny last updated on 01/Jun/21

$$conditionsx\ne 0y\ne 0xy\ne 0$$$$\frac{1}{x}=t\frac{1}{y}=u$$$$t+u+tu=\frac{1}{4}$$$$t(1+u)=\frac{1}{4}-u$$$$t=-\frac{u-\frac{1}{4}}{1+u}$$$$-\frac{u-\frac{1}{4}}{1+u}+u-\left(\frac{u-\frac{1}{4}}{1+u}\right)\times u=\frac{1}{4}$$$$-\frac{4u-1}{1+u}+u-\left(\frac{4u-4}{1+u}\right)\times u=1$$$$-\frac{4u-1}{1+u}+\frac{u(1+u)}{1+u}-\left(\frac{4u^2-4u}{1+u}\right)=1$$$$-4u+1+u(1+u)-(4u^2-4u)=1+u$$$$-4u+1+u+u^2-4u^2+4u=1+u$$$$-3u^2=0u=0$$$$t=-\frac{u-\frac{1}{4}}{1+u}$$$$t=-\frac{-\frac{1}{4}}{1}t=\frac{1}{4}=\frac{1}{x}$$$$$$$$$$$$$$

Answered by ArielVyny last updated on 01/Jun/21

$$itispossible?$$

Answered by ArielVyny last updated on 01/Jun/21

$$ithinkisnotpossible$$

Answered by ajfour last updated on 01/Jun/21

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}=\frac{1}{4}$$$$\Rightarrow 4x+4y+4=xy$$$$(x-4)(y-4)=20$$$$=1\times 20=2\times 10=4\times 5$$$$\Rightarrow (x,y)=(5,24),(6,14),(8,9)$$

Commented by mnjuly1970 last updated on 01/Jun/21

$$thsnksalot...$$$$(2,-6),(6,-2),(-16,3),(3,-16)$$$$forexample:$$$$(2,-6)$$$$\frac{1}{2}-\frac{1}{6}-\frac{1}{12}=\frac{6-2-1}{12}=\frac{1}{4}..✓$$$$$$

Answered by mnjuly1970 last updated on 01/Jun/21

$$solution:$$$$\frac{y+x+1}{xy}=\frac{1}{4}$$$$4x+4y+4-xy=0$$$$x(4-y)=-4y-4$$$$x=\frac{-4y-4}{4-y}=\frac{4(y+1)}{y-4}=4.\frac{y-4+5}{y-4}$$$$=4+\frac{20}{y-4}....x,y\in \mathbb{Z}$$$$y-4=\pm 1,y-4=\pm 2,y-4=4$$$$y-4=5,y-4=\pm 10$$$$y-4=\pm 20$$$$,......easy...$$$$$$

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