Solve the equation $\frac{p}{q}-\frac{4}{r+1}=1$ in prime numbers.
We can rewrite the equation in the form

$$
\begin{gathered}
\frac{p r+p-4 q}{q(r+1)}=1 \Rightarrow p r+p-4 q=q r+q \\
p r-q r=5 q-p \Rightarrow r(p-q)=5 q-p
\end{gathered}
$$

It follows that \( p \neq q \) and

$$
\begin{gathered}
r=\frac{5 q-p}{p-q}=\frac{4 q+q-p}{p-q} \\
r=\frac{4 q}{p-q}-1
\end{gathered}
$$

As \( p \) is prime, \( p-q \neq q \), \( p-q \neq 2 q \), \( p-q \neq 4 q \).

We have \( p-q=1 \) or \( p-q=2 \) or \( p-q=4 \)

i) If \( p-q=1 \) then

$$
q=2, p=3, r=7
$$

ii) If \( p-q=2 \) then \( p=q+2, r=2 q-1 \)

If \( q=1(\bmod 3) \) then \( q+2 \equiv 0(\bmod 3) \)

$$
q+2=3 \Rightarrow q=1
$$

contradiction.

If \( q \equiv-1(\bmod 3) \) then \( r \equiv-2-1 \equiv 0(\bmod 3) \)

$$
\begin{gathered}
r=3 \\
r=2 q-1=3 \\
q=2 \\
p=4
\end{gathered}
$$

contradiction.

Hence \( q=3, p=5, r=5 \).

iii) If \( p-q=4 \) then \( p=q+4 \).

\( r=q-1 \)

Hence \( q=3, p=7, r=2 \).