Venerable Soul!
Find ∑ ( x + y ) if ( x , y ) indicates the pairs that are the solutions of the equation- ( x 2 + 1 ) ( y 2 + 1 ) + 2 ( x − y ) ( 1 − x y ) = 4 ( 1 + x y )
x , y ∈ Z
Note-This problem is not mine. ∑ ( x + y ) indicates sum of all pairs satisfying the equality
The answer is 0.
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3 solutions
You meant x , y ∈ Z , so please add it.
Notice that 2 ( x − y ) ( 1 − x y ) = 2 x ( y 2 + 1 ) − 2 y ( x 2 + 1 ) .
( x 2 + 1 ) ( y 2 + 1 ) ⟺ ( x 2 + 1 ) ( y 2 + 1 ) + 2 ( x − y ) ( 1 − x y ) + 2 x ( y 2 + 1 ) − 2 y ( x 2 + 1 ) = 4 ( 1 + x y ) = 4 ( 1 + x y )
Remember that a b + m a + n b + m n = ( a + n ) ( b + m ) . Thus:
⟺ ( x 2 + 1 + 2 x ) ( y 2 + 1 − 2 y ) ⟺ ( x + 1 ) 2 ( y − 1 ) 2 ⟺ ( x + 1 ) ( y − 1 ) = 4 = 4 = ± 2 ⟺ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ { x + 1 y − 1 = 2 = 1 or { x + 1 y − 1 = − 2 = − 1 or { x + 1 y − 1 = − 2 = 1 or { x + 1 y − 1 = 2 = − 1 or { x + 1 y − 1 = 1 = 2 or { x + 1 y − 1 = − 1 = − 2 or { x + 1 y − 1 = − 1 = 2 or { x + 1 y − 1 = 1 = − 2 ⟺ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ { x y = 1 = 2 or { x y = − 3 = 0 or { x y = − 3 = 2 or { x y = 1 = 0 or { x y = 0 = 3 or { x y = − 2 = − 1 or { x y = − 2 = 3 or { x y = 0 = − 1 ⟺ i = 0 ∑ 7 ( x i + y i ) = 0
Okay! I shall add it in! Voted you up for the pains you took to write so much in LaTex!
Was this problem too easy a Level 5?
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It should be a Level 4, I think! After simplifying, it is easy to find the solutions.
( x 2 + 1 ) ( y 2 + 1 ) + 2 ( x − y ) ( 1 − x y ) = 4 + 4 x y
x 2 y 2 + x 2 + y 2 + 1 − 4 x y + 2 ( x − y ) ( 1 − x y ) = 4
x 2 y 2 + 1 − 2 x y + x 2 + y 2 − 2 x y + 2 ( x − y ) ( 1 − x y ) = 4
( x − y ) 2 + ( 1 − x y ) 2 + 2 ( x − y ) ( 1 − x y ) = 4
( x − y + x y − 1 ) 2 = 4
x − y + x y − 1 = 2
x − y − x y + 1 = 2
x − y − x y = 1
Let us assume that x = 0. Solving quadratic in y with x = 0, we get ........
y = -1.. OR.. y = 3. Solution (0, -1) , (0, 3).
In same way for y = 0, we get..
( -3, 0), (1, 0). Adding them all we get 0.
No tedious math needed:
Suppose that ( a , b ) is a solution. Then, it is easy to see that ( − b , − a ) is also a solution. Thus, the sum of all the solutions must be 0 .