
6. Find all pairs $(x, y)$ of real numbers such that

$$
16^{x^{2}+y}+16^{x+y^{2}}=1
$$


Solution: Observe that

$$
x^{2}+y+x+y^{2}+\frac{1}{2}=\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2} \geq 0
$$

This shows that $x^{2}+y+x+y^{2} \geq(-1 / 2)$. Hence we have

$$
\begin{aligned}
1=16^{x^{2}+y}+16^{x+y^{2}} & \geq 2\left(16^{x^{2}+y} \cdot 16^{x+y^{2}}\right)^{1 / 2}, \quad \text { (by AM-GM inequality) } \\
& =2\left(16^{x^{2}+y+x+y^{2}}\right)^{1 / 2} \\
& \geq 2(16)^{-1 / 4}=1
\end{aligned}
$$

Thus equality holds every where. We conclude that

$$
\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=0
$$

This shows that $(x, y)=(-1 / 2,-1 / 2)$ is the only solution, as can easily be verified.

