16(!). Solve the equation \(x+y=x y\) in integers.
16. Solution. First method. Noting that \(y \neq 1\), we get:

\[
x=\frac{y}{y-1}=\frac{(y-1)+1}{y-1}=1+\frac{1}{y-1}
\]

But the fraction \(\frac{1}{y-1}\) can take integer values only when \(y-1=1\) or \(y-1=-1\). In the first case, \(y=2, x=2\). In the second case, \(x=y=0\).

Second method. The simplicity of the answers might suggest another approach. For example, we can try to verify the correctness of the two solutions mentioned above and show that no other solutions are possible.

Assuming the cases \(x y=0, x=y=2\) are excluded.

Dividing the given equation term by term by \(x y\), we get:

\[
\frac{1}{x}+\frac{1}{y}=1
\]

For integers \(x\) and \(y\) not equal to 1, \(\frac{1}{x}<0.5\) and \(\frac{1}{y}<0.5\), so the sum of these fractions is less than 1. It is also clear that \(x \neq 1, y \neq 1\). Therefore, no other solutions are possible.

Third method. From the given equation, it follows that \(x y-x- y+1=1\) or after grouping terms on the left side of the equation \((x-1)(y-1)=1\).

The product of integers \(x-1\) and \(y-1\) can be equal to 1 if each of these numbers is 1 or -1. This conclusion allows us to find the values of the unknowns.