Solve the following equation if $x$ and $y$ are real numbers:

$$
\left(16 x^{2}+1\right)\left(y^{2}+1\right)=16 x y .
$$
I. solution. The left side of the equation is positive, so neither $x$ nor $y$ can be zero. Dividing both sides by $4 x y$, we arrive at the equivalent equation

$$
\left(4 x+\frac{1}{4 x}\right)\left(y+\frac{1}{y}\right)=4
$$

It is known that for any real number $a$, $\left|a+\frac{1}{a}\right| \geq 2$, and equality holds if and only if $a= \pm 1$. Therefore, on the left side of (1), the absolute value of each factor is at least 2, so for equality, $|4 x|=|y|=1$ is necessary, as well as the requirement that the signs of the two factors on the left side of (1) be the same.

From this, two solutions arise: from $4 x=y=1$, we get $x=\frac{1}{4}$ and $y=1$, and from $4 x=y=-1$, we get $x=-\frac{1}{4}$ and $y=-1$. It can be seen that both pairs of numbers are solutions to the problem.

II. solution. Performing the multiplication on the left side of the equation and rearranging, we write the resulting two-variable polynomial in terms of powers of $x$:

$$
16\left(y^{2}+1\right) x^{2}-16 y x+\left(y^{2}+1\right)=0
$$

Since $16\left(y^{2}+1\right)$ is positive, the left side of (2) is zero precisely when

$$
x^{2}-\frac{y}{y^{2}+1} x+\frac{1}{16}=0
$$

For a given $y$, there exist 0, 1, or 2 $x$ such that the pair $(x, y)$ is a solution to the equation, depending on whether the discriminant $D(y)$ of the parametric equation (3) is negative, zero, or positive. $D(y)$ can be factored as a difference of squares:

$$
\begin{aligned}
D(y) & =\left(\frac{y}{y^{2}+1}\right)^{2}-\frac{1}{4}=\left(\frac{y}{y^{2}+1}+\frac{1}{2}\right)\left(\frac{y}{y^{2}+1}-\frac{1}{2}\right)= \\
& =\frac{2 y+y^{2}+1}{2\left(y^{2}+1\right)} \cdot \frac{2 y-\left(y^{2}+1\right)}{2\left(y^{2}+1\right)}
\end{aligned}
$$

thus

$$
D(y)=\frac{(y+1)^{2}}{2\left(y^{2}+1\right)} \cdot\left(-\frac{(y-1)^{2}}{2\left(y^{2}+1\right)}\right)=-\left(\frac{y^{2}-1}{2\left(y^{2}+1\right)}\right)^{2}
$$

We find that $D(y)$ is the negative of a perfect square, and thus cannot be positive. We obtain solutions for $y$ precisely when $D(y)=0$, that is,

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

which holds precisely when $y= \pm 1$. If $y=1$, then according to (3),

$$
x^{2}-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^{2}=0, \quad x=\frac{1}{4}
$$

while if $y=-1$,

$$
x^{2}+\frac{x}{2}+\frac{1}{16}=\left(x+\frac{1}{4}\right)^{2}=0 \quad x=-\frac{1}{4}
$$

Thus, the equation has two solutions: the pairs $\left(\frac{1}{4} ; 1\right)$ and $\left(-\frac{1}{4} ;-1\right)$.

Remarks. 1. Using complex numbers, the left side of (3) can be rewritten as follows:

Let $f(y)=\frac{y}{y^{2}+1}, g(y)=\frac{y^{2}-1}{2\left(y^{2}+1\right)}$; then (3) holds precisely when

$$
x_{1,2}=\frac{f(y) \pm \sqrt{-g^{2}(y)}}{2}=\frac{f(y)+i g(y)}{2} \text { where } i^{2}=-1
$$

From this, we find that the left side of (3) can be written in the following "root factor" form:

$$
\left(x-\frac{f(y)+i g(y)}{2}\right)\left(x-\frac{f(y)-i g(y)}{2}\right)
$$

Performing the multiplication, we get

$$
x^{2}-f(y) x+\frac{f^{2}(y)}{4}+\frac{g^{2}(y)}{2}=\left(x-\frac{f(y)}{2}\right)^{2}+\left(\frac{g(y)}{2}\right)^{2}
$$

that is, the left side of (3) is a sum of two squares:

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

2. The two-variable polynomial (2) can also be factored into a sum of two squares in a different way; it is easily verified that

$$
\left(16 x^{2}+1\right)\left(y^{2}+1\right)-16 x y=(4 x y-1)^{2}+(4 x-y)^{2}
$$

From this form, we can also arrive at the solution, since the sum of the squares of two real numbers is zero precisely when both numbers are zero.