8.6. Determine all ordered pairs of prime numbers $(m, n)$ such that the numbers $2 m+n$, $m+2 n$ and $m+n-28$ are prime numbers.
Solution. Since the numbers $2m+n, m+2n$ are prime, it follows that the prime numbers $m$ and $n$ are both odd. Then the numbers $2m+n$ and $m+2n$ are odd prime numbers, and their sum $(2m+n)+(m+2n)=3 \cdot(m+n)$ is an even number. Since 3 is an odd number, it follows that the number $m+n$ is even. Then the number $m+n-28$ is even and prime, i.e., $m+n-28=2 \Leftrightarrow m+n=30$. By assigning values, we find the following 6 ordered pairs of odd prime numbers with a sum equal to 30: $(7,23),(11,19),(13,17),(17,13),(19,11),(23,7)$. The values of the numbers $2m+n$ and $m+2n$ for each of these 6 pairs are inserted in the table:

| $(\mathbf{m}, \mathbf{n})$ | $(7,23)$ | $(11,19)$ | $(13,17)$ | $(17,13)$ | $(19,11)$ | $(23,7)$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathbf{2m}+\mathbf{n}$ | 37 | 41 | 43 | 47 | 49 | 53 |
| $\mathbf{m}+\mathbf{2n}$ | 53 | 49 | 47 | 41 | 41 | 37 |

From the table, it follows that the pairs $(11,19)$ and $(19,11)$ do not satisfy the problem statement. The other four remaining pairs $(7,23),(13,17),(17,13),(23,7)$ represent all the ordered pairs of prime numbers that satisfy the problem statement.