Mathematical Reflections 2013, Issue 5 - Problem J279

Problem:
Find all triples $(p, q, r)$ of primes such that $pqr = p + q + r + 2000$.

Proposed by Titu Andreescu.

Solution:
Assume without loss of generality that $p \leq q \leq r$. The given equality can be rewritten as
\begin{equation}\label{first-eq}
(rq-1)(p-1)+(r-1)(q-1)=2002.                 (1)
\end{equation}
If $p$ is an odd prime, then $q$ and $r$ are odd prime also, but this means that the LHS is divisible by $4$ and the RHS is not divisible by $4$, a contradiction. Thus, $p=2$ and equation $(1)$ becomes $$(2q-1)(2r-1)=4005=3^2\cdot5\cdot89.$$ Since $2q-1 \leq 2r-1$, then $(2q-1)^2 \leq 4005$, i.e. $2q-1 \leq 63$. This means that $2q-1 \in \{1,3,5,9,15,45\}$. Clearly, $2q-1 \neq 1,15$, therefore we have the four systems of equations
$$\begin{array}{lll} 2q-1&=&3 \\ 2r-1&=&1335, \end{array} \qquad \begin{array}{lll} 2q-1&=&5 \\ 2r-1&=&801, \end{array} \qquad \begin{array}{lll} 2q-1&=&9 \\ 2r-1&=&445, \end{array} \qquad \begin{array}{lll} 2q-1&=&45 \\ 2r-1&=&89. \end{array}$$
It's easy to see that the first and the last system have no solution in primes, and the other two systems give $q=3, r=401$ and $q=5, r=223$. Therefore, $(2,3,401)$ and $(2,5,223)$ are two solutions to the given problem and by symmetry all the solutions are $$(2,3,401),(2,401,3),(3,2,401),(3,401,2),(401,2,3),(401,3,2),$$ $$(2,5,223),(2,223,5),(5,2,223),(5,223,2),(223,2,5),(223,5,2).$$