Number of real ordered pair $(x, y)$ satisfying $x^{2} + 1 = y$ and $y^{2} + 1 = x$ is

0

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Solution

The correct option is A $0$
$x^{2} + 1 = y$ and $y^{2} + 1 = x$
Subtracting $x^{2} - y^{2} = y - x$
$\Rightarrow (x - y) (x + y + 1) = 0$
$\Rightarrow x = y$ or $x + y + 1 = 0$
If $x = y$ then $x^{2} + 1 = y \Rightarrow x^{2} - x + 1 = 0$ which has no real roots.
If $x + y + 1 = 0$ then $x + y = - 1$
Adding given equations, $x^{2} + y^{2} + 2 = x + y$
$\Rightarrow x^{2} + y^{2} + 2 = - 1 \Rightarrow x^{2} + y^{2} + 3 = 0$ which is not possible.

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