Find the number of real ordered pair(s) (x,y) for which : $16^{x^{2} + y} + 16^{x + y^{2}} = 1$

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answer is B.

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Detailed Solution

$\begin{array}{l} {(2^{2 x^{2} + 2 y} - 2^{2 x + 2 y^{2}})}^{2} = 1 - 2^{2 x^{2} + 2 y^{2} + 2 x + 2 y + 1} \geq 0 \\ \Rightarrow {(x + \frac{1}{2})}^{2} + {(y + \frac{1}{2})}^{2} \leq 0 \end{array}$

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