Very very beautiful

Algebra Level 5

Consider the equation ( x + y ) 2 = ( x + 1 ) ( y − 1 ) (x+y)^2=(x+1)(y-1)

If all the possible real ordered pairs ( x , y ) (x,y) that satisfy the equation above are ( x 1 , y 1 ) , ( x 2 , y 2 ) , . . . , ( x n , y n ) (x_1,y_1),(x_2,y_2),...,(x_n,y_n) then find the value of x 1 2 + . . . + x n 2 + y 1 2 + . . . + y n 2 x^2_1+...+x^2_n+y^2_1+...+y^2_n .

  • If you think there are infinite solutions then answer 777 and if you think no real solutions answer 666.
  • Ordered pair means (11,12),(12,11) are considered different.

This is a part of the set Beautiful..It is!!


The answer is 2.

Ravi Dwivedi by Ravi Dwivedi , 24, India

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1 solution

Ravi Dwivedi
Jul 12, 2015

It is this solution why this problem is indeed beautiful

Put x + 1 = X , y − 1 = Y x+1=X, y-1=Y

The equation becomes

( X + Y ) 2 = X Y (X+Y)^2=XY\\ 1 2 ( X 2 + Y 2 + ( X + Y ) 2 ) = 0 \frac{1}{2}(X^2+Y^2+(X+Y)^2)=0\\

All terms on LHS are perfect squares and 0 on RHS which forces

X = 0 , Y = 0 ⟹ x = − 1 , y = 1 X=0,Y=0 \implies x=-1,y=1

Therefore ( − 1 , 1 ) (-1,1) is the only ordered pair which satisfies the given solution and hence the answer is ( − 1 ) 2 + 1 2 = 2 (-1)^2+1^2=\boxed{2}

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Moderator note:

One can also directly apply AM-GM to your solution, and conclude that we are in the equality case which gives us x + 1 = y − 1 = 0 x+ 1 = y - 1 = 0 .

In a sense, it is equivalent to your approach, though it doesn't require us to see the transformation of variables.

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