Solve the following equation:

$$
16^{x^{2}+y}+16^{y^{2}+x}=1
$$
Both terms on the right-hand side are positive, so we can apply the arithmetic and geometric mean inequality:

$$
\begin{aligned}
& 1=16^{x^{2}+y}+16^{y^{2}+x} \geq 2 \cdot \sqrt{16^{x^{2}+y} \cdot 16^{y^{2}+x}}=2 \cdot \sqrt{16^{x^{2}+y+y^{2}+x}}= \\
& \quad=2 \cdot 4^{x^{2}+y+y^{2}+x}=A \\
& x^{2}+y+y^{2}+x=x^{2}+x+\frac{1}{4}+y^{2}+y+\frac{1}{4}-\frac{1}{2}=\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}-\frac{1}{2}
\end{aligned}
$$

Using this:

$$
A=2 \cdot 4^{\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}-\frac{1}{2}} \geq 2 \cdot 4^{-\frac{1}{2}}=1
$$

It is clear that equality holds if and only if

$\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=0$, which means $x=-\frac{1}{2}$ and $y=-\frac{1}{2}$, and this is indeed a solution to the equation:

$$
16^{\frac{1}{4}-\frac{1}{2}}+16^{\frac{1}{4}-\frac{1}{2}}=2 \cdot 16^{-\frac{1}{4}}=\frac{2}{\sqrt[4]{16}}=\frac{2}{2}=1
$$