Is there a natural number for which all the sums and differences of its factor pairs are prime?

It seems the following.

There is a following recursive way to search such the integer $N$.

$N=1\cdot k_1$. Since all of numbers $k_1\pm 1$ are prime, one of these numbers is odd. So we have

$N=1\cdot 2\cdot k_2$. Since all of numbers $2k_2\pm 1$, $k_2\pm 2$ are prime, if $k_2$ is not divisible by $3$, then one of these numbers is divisible by $3$, so it equals $3$. This case is considered separately, and in the rest of the cases we have

$N=1\cdot 2\cdot 3\cdot k_3$. Since all of numbers $6k_3\pm 1$, $3k_3\pm 2$, $2k_3\pm 3$, $k_3\pm 6$ are prime, if $k_2$ is not divisible by $7$, then one of these numbers is divisible by $7$, so it equals $7$. This case is considered separately, and in the rest of the cases we have

$N=1\cdot 2\cdot 3\cdot 7\cdot k_4$. Since all of numbers $42k_4\pm 1$, $21k_4\pm 2$ are prime, if $k_4$ is not divisible by $5$, then one of these numbers is divisible by $5$, so it equals $5$. This case is considered separately, and in the rest of the cases we have

$N=1\cdot 2\cdot 3\cdot 7\cdot 5 \cdot k_5$, and so forth...