#2015_8 I know that you know it

Number Theory Level 4

( x 2 + 1 ) ( y 2 + 1 ) + 2 ( x − y ) ( 1 − x y ) = 4 ( 1 + x y ) ({ x }^{ 2 }+1)({ y }^{ 2 }+1)+2(x-y)(1-xy)=4(1+xy)

What is number of integral solutions for the equation above?

More questions?? .


The answer is 8.

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Anandhu Raj by Anandhu Raj , 23, India

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1 solution

Anandhu Raj
Mar 14, 2015

Write the equation ( x 2 + 1 ) ( y 2 + 1 ) + 2 ( x − y ) ( 1 − x y ) = 4 ( 1 + x y ) ({ x }^{ 2 }+1)({ y }^{ 2 }+1)+2(x-y)(1-xy)=4(1+xy) in the form,

x 2 y 2 + y 2 + x 2 + 1 + 2 ( x − y ) ( 1 − x y ) = 4 + 4 x y { x }^{ 2 }{ y }^{ 2 }+{ y }^{ 2 }+{ x }^{ 2 }+1+2(x-y)(1-xy)=4+4xy

⇒ x 2 y 2 − 2 x y + 1 + y 2 + x 2 − 2 x y + 2 ( x − y ) ( 1 − x y ) = 4 \Rightarrow { x }^{ 2 }{ y }^{ 2 }-2xy+1+{ y }^{ 2 }+{ x }^{ 2 }-2xy+2(x-y)(1-xy)=4

⇒ ( x y − 1 ) 2 + ( x − y ) 2 − 2 ( x − y ) ( x y − 1 ) = 4 \Rightarrow { (xy-1) }^{ 2 }+({ x-y })^{ 2 }-2(x-y)(xy-1)=4

⇒ [ x y − 1 − ( x − y ) ] 2 = 4 \Rightarrow { [xy-1-(x-y) }]^{ 2 }=4

⇒ x y − 1 − ( x − y ) = ± 2 \Rightarrow { xy-1-(x-y) }=\pm 2 \

⇒ ( x + 1 ) ( y − 1 ) = ± 2 \Rightarrow (x+1)(y-1)=\pm 2

CASE I: ( x + 1 ) ( y − 1 ) = 2 (x+1)(y-1)=2 , we obtain system of equations,

{ x + 1 = 2 y − 1 = 1 } { x + 1 = − 2 y − 1 = − 1 } { x + 1 = 1 y − 1 = 2 } { x + 1 = − 1 y − 1 = − 2 } \begin{Bmatrix} x+1=2 \\ y-1=1 \end{Bmatrix} \quad \quad \quad \quad \quad \quad \begin{Bmatrix} x+1=-2 \\ y-1=-1 \end{Bmatrix} \\ \begin{Bmatrix} x+1=1 \\ y-1=2 \end{Bmatrix}\quad \quad \quad \quad \quad \quad \begin{Bmatrix} x+1=-1 \\ y-1=-2 \end{Bmatrix}

yielding solutions (1,2),(-3,0),(0,3),(-2,-1)

CASE II: ( x + 1 ) ( y − 1 ) = − 2 (x+1)(y-1)=-2 , we obtain system of equations,

{ x + 1 = 2 y − 1 = − 1 } { x + 1 = − 2 y − 1 = 1 } { x + 1 = 1 y − 1 = − 2 } { x + 1 = − 1 y − 1 = 2 } \begin{Bmatrix} x+1=2 \\ y-1=-1 \end{Bmatrix} \quad \quad \quad \quad \quad \quad\begin{Bmatrix} x+1=-2 \\ y-1=1 \end{Bmatrix} \\ \begin{Bmatrix} x+1=1 \\ y-1=-2 \end{Bmatrix} \quad \quad \quad \quad \quad \quad \begin{Bmatrix} x+1=-1 \\ y-1=2 \end{Bmatrix}

yielding solutions (1,0),(-3,2),(0,-1),(-2,3)

Thus there are 8 \boxed{8} integral solutions.

6 Helpful 0 Interesting 0 Brilliant 0 Confused

{ x + 1 = 2 y − 1 = 1 } { x + 1 = − 2 y − 1 = − 1 } [ x + 1 = 1 y − 1 = 2 ] [ x + 1 = − 1 y − 1 = − 2 ] \begin{Bmatrix} x+1=2 \\ y-1=1 \end{Bmatrix} \quad \quad \quad \quad \quad \quad \begin{Bmatrix} x+1=-2 \\ y-1=-1 \end{Bmatrix} \\ \begin{bmatrix} x+1=1 \\ y-1=2 \end{bmatrix} \quad \quad \quad \quad \quad \quad \begin{bmatrix} x+1=-1 \\ y-1=-2 \end{bmatrix}

Don't use \left { and \right in the code for matrixs. Use {bmatrix} for square brackets and {Bmatrix} for curly brackets.

I can't edit your Latex since I'm not a mod. So to fix your code, delete all "\left {" and "\right". You can put the brackets you want by adding a "b" or "B" to {matrix} (both begin and end).

Siddhartha Srivastava - 6 years, 3 months ago

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@Siddhartha Srivastava Thanks a lot!:)

Anandhu Raj - 6 years, 3 months ago

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