Is there a natural number for which all the sums and differences of its factor pairs are prime? The 8 factor pairs of e.g. 462 are
$((1, 462), (2, 231), (3, 154), (6, 77), (7, 66), (11, 42), (14, 33), (21, 22))$.
Of the 16 non-negative integers which are the sums and differences of these pairs (such as $462+1=463$, $462-1=461$, $21+22=43$, and $22-21=1$), 15 of them are primes. (The only non-prime is $22-21=1$.)
Is there an integer for which all the $2n$ sums and differences of its $n$ factor pairs are primes?
Obviously any such integer needs be the even number between a twin prime pair, and needs to be $ = 2$ (mod 4), but that's all I figured out.

Amongst integers less than $10^7$, 462 seems to have the uniquely highest fraction of prime sums/differences of factor pairs, with $15/16$. If there is no integer with a fraction of 1 (the question above), can there be any with a higher fraction than 15/16?
 A: 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... 
