## Task A-2.3.

Determine all ordered triples $(m, n, p)$, where $m$ and $n$ are natural numbers, and $p$ is a prime number, for which

$$
(2 m+3)(4 n+1)=p m n
$$
## First Solution.

Notice that $m \mid (2m+3)(4n+1)$. Since $D(m, 2m+3) = D(m, 3)$, we consider the cases $D(m, 3) = 1$ and $D(m, 3) = 3$ (i.e., $3 \mid m$).

First, assume that $D(m, 3) = 1$. Therefore, $m$ must divide $4n+1$. Similarly, since $n \mid (2m+3)(4n+1)$ and $n$ and $4n+1$ are relatively prime, we have $n \mid 2m+3$.

The initial equation can be written as

$$
\frac{2m+3}{n} \cdot \frac{4n+1}{m} = p
$$

where the left side is the product of two natural numbers. Since $p$ is a prime number, we have two cases: one of them is 1, and the other is $p$ or vice versa.

In the first case, we have $2m+3 = np$ and $4n+1 = m$. Substituting, we get $2(4n+1) + 3 = np$, or $(p-8)n = 5$. It cannot be that $p-8 = 1$ because $p$ would not be a prime number, so the only solution is $n = 1, p-8 = 5$, i.e., $p = 13$ and $m = 5$.

In the second case, we have $2m+3 = n$ and $4n+1 = mp$. Substituting, we get $4(2m+3) + 1 = mp$, or $m(p-8) = 13$. From $p-8 = 1$ and $p-8 = 13$, we would get $p = 9$ or $p = 21$, which are not prime numbers, so this case has no solutions.

Now, assume that $3 \mid m$. Thus, $m$ can be written as $m = 3k, k \in \mathbb{N}$. Substituting into the initial equation and simplifying, we get $(2k+1)(4n+1) = pkn$.

Since $D(2k+1, k) = 1$ and $D(4n+1, n) = 1$, we similarly conclude that $k \mid 4n+1$ and $n \mid 2k+1$.

Therefore, the initial equation can be written as

$$
\frac{2k+1}{n} \cdot \frac{4n+1}{k} = p
$$

where the left side is the product of two natural numbers, and we consider two cases: one of them is 1, and the other is $p$ or vice versa.

In the first case, we have $2k+1 = np$ and $4n+1 = k$. Substituting, we get $2(4n+1) + 1 = np$, or $n(p-8) = 3$. The only solution is $n = 1, p-8 = 3$, i.e., $p = 11, k = 5$, $m = 15$.

In the second case, we have $2k+1 = n$ and $4n+1 = kp$, or $4(2k+1) + 1 = kp, k(p-8) = 5$. This gives $k = 1, p-8 = 5$, i.e., $p = 13, m = 3, n = 3$.

Therefore, all solutions are $(m, n, p) = (5, 1, 13), (15, 1, 11)$, and $(3, 3, 13)$.