Solve the following equation if $x, y, z$, and $v$ are positive integers.

$$
x+\frac{1}{y+\frac{1}{z+\frac{1}{v}}}=\frac{101}{91} .
$$
Since $x, y, z$ and $v$ are positive integers, we have

$$
0<\frac{1}{y+\frac{1}{z+\frac{1}{v}}}<1
$$

Based on this, only $x=1$ is possible. In this case, we arrive at the following equation:

$$
y+\frac{1}{z+\frac{1}{v}}=\frac{91}{10}=9+\frac{1}{10}
$$

Since $y, z, v$ are positive integers, we have

$$
0<\frac{1}{z+\frac{1}{v}}<1
$$

which implies $y=9$. Using this, we get the equation

$$
\frac{1}{v}=10-z
$$

The right-hand side of this equation is an integer, so the left-hand side must also be an integer. Since $v$ is a positive integer, its reciprocal can only be an integer if $v=1$. This implies $z=9$.

Therefore, we have proven that the only solution to the equation is the quadruple $x=1 ; y=9 ; z=9 ; v=1$.

Based on the work of Turchányi Judit (Budapest, Berzsenyi D. Gymnasium, 1st year). Note: The left-hand side of the given equation is called the continued fraction expansion of the right-hand side.