Find all positive integers $p$ such that $p, p+4$, and $p+8$ are all prime.
If $p=3$, then $p+4=7$ and $p+8=11$, both prime. If $p \neq 3$, then $p$ is not a multiple of 3 and is therefore of one of the forms $3 k+1,3 k+2(k \geq 0)$ If $p=3 k+1$, then $p+8=3 k+9=3(k+3)$, which is not prime since $k+3>1$. If $p=3 k+2$, then $p+4=3 k+6=3(k+2)$, which is not prime since $k+2>1$. Thus $p=3$ is the only solution with all three numbers prime.