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Question Number 61650 by maxmathsup by imad last updated on 05/Jun/19

$$solveinside{N}^2(x+1)(y+2)=2xy$$

Commented by kaivan.ahmadi last updated on 06/Jun/19

$$xy+2x+y+2-2xy=0$$$$xy=2x+y+2$$$$letx=m\in N$$$$my=2m+y+2\Rightarrow (m-1)y=2m+2\Rightarrow$$$$y=\frac{2m+2}{m-1}=2\left(\frac{m+1}{m-1}\right)=2(1+\frac{2}{m-1})=$$$$2+\frac{4}{m-1}$$$$m-1⩽4\Rightarrow m⩽5$$$$m=1isnottrue$$$$m=2\Rightarrow y=6$$$$m=3\Rightarrow y=4$$$$m=4isnottrue$$$$m=5\Rightarrow y=3$$$$so$$$$(2,6)(3,4)(5,3)areanswer$$$$$$$$$$

Commented by maxmathsup by imad last updated on 06/Jun/19

$$(x+1)(y+2)=2xy\Rightarrow xy+2x+y+2=2xy\Rightarrow xy+2x+y+2-2xy=0\Rightarrow$$$$2x+y+2-xy=0letconsidercongruencemodulo2\left(sovinginZ/2Z\right)\Rightarrow$$$$\stackrel{-}{2}\stackrel{-}{x}+\stackrel{-}{y}+\stackrel{-}{2}-\stackrel{-}{x}\stackrel{-}{y}=0\Rightarrow \stackrel{-}{y}(\stackrel{-}{1}-\stackrel{-}{x})=0\Rightarrow \stackrel{-}{y}=0or\stackrel{-}{x}=\stackrel{-}{1}$$$$y=2n\Rightarrow (x+1)(2n+2)=2\left(2n\right)x\Rightarrow 2nx+2x+2n+2=4nx\Rightarrow$$$$(2n+2-4n)x+2n+2=0\Rightarrow (2-2n)x=-2n-2\Rightarrow (n-1)x=n+1\Rightarrow x=\frac{n+1}{n-1}$$$$x\in N\Rightarrow \frac{n-1+2}{n-1}\in N\Rightarrow 1+\frac{2}{n-1}\in N\Rightarrow n-1/2\Rightarrow n=2orn=3\Rightarrow x=3orx=2$$$$\Rightarrow y=4ory=6sothecouplesare(3,4)and(2,6)$$$$\stackrel{-}{x}=\stackrel{-}{1}\Rightarrow x=2n+1\left(e\right)\Rightarrow (2n+2)(y+2)=2(2n+1)y\Rightarrow$$$$(n+1)(y+2)=(2n+1)y\Rightarrow (n+1)y+2n+2=(2n+1)y\Rightarrow$$$$(n+1)y=2(n+1)\Rightarrow y=2\Rightarrow 4(x+1)=4x\Rightarrow 4=0impossibe$$

Commented by maxmathsup by imad last updated on 06/Jun/19

$$(5,3)isalsoanswer.idontgetit...!$$

Commented by maxmathsup by imad last updated on 06/Jun/19

$$thankssirAhmadi.$$

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