Does symmetry works always ?

Algebra Level pending

( x 2 + 1 ) ( y 2 + 1 ) + 25 = 10 x + 10 y \displaystyle{({ x }^{ 2 }+1)({ y }^{ 2 }+1)+25=10x+10y}

Find value of x 2 + y 2 \displaystyle{\\ { x }^{ 2 }+{ y }^{ 2 }}

This is not original


The answer is 23.

Karan Shekhawat by Karan Shekhawat , 25, India

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1 solution

Tijmen Veltman
Apr 15, 2015

We expand to:

x 2 y 2 + x 2 + y 2 + 1 + 25 − 10 ( x + y ) = 0 x^2y^2+x^2+y^2+1+25-10(x+y)=0

( x y ) 2 + 1 − 2 ( x y ) + 2 ( x y ) + x 2 + y 2 − 10 ( x + y ) + 25 = 0 (xy)^2+1-2(xy)+2(xy)+x^2+y^2-10(x+y)+25=0

( x y − 1 ) 2 + ( x + y ) 2 − 10 ( x + y ) + 25 = 0 (xy-1)^2+(x+y)^2-10(x+y)+25=0

( x y − 1 ) 2 + ( x + y − 5 ) 2 = 0. (xy-1)^2+(x+y-5)^2=0.

This only holds if x y = 1 xy=1 and x + y = 5 x+y=5 , hence x 2 + y 2 = ( x + y ) 2 − 2 x y = 25 − 2 = 23 x^2+y^2=(x+y)^2-2xy=25-2=\boxed{23} .

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