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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?

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  • $\begingroup$ Noting that twin primes (except 3,5) are of the form (6n-1,6n+1), you can say more strongly that such a number would have to be $6\pmod {12}$. $\endgroup$ – Milo Brandt Apr 3 '15 at 3:34
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    $\begingroup$ And such number has to be square-free. $\endgroup$ – punctured dusk Apr 3 '15 at 11:06
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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...

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  • $\begingroup$ If $p_1,\ldots,p_{n-1}\mid N$ and we want to prove that $p_n\mid N$, this method gives $2^{\Omega(p_n^2)}$ numbers none of which can be divisible by $p_n$ but which I expect to be more or less uniformly distributed modulo $p_n$. Seems likely that we will be able to show that for sufficiently large $n$, $p_1,\ldots,p_n\mid N$ implies $p_{n+1}\mid N$. This way it becomes an analytic number theory question. $\endgroup$ – punctured dusk Apr 3 '15 at 11:14
  • $\begingroup$ @barto I have a hope that it can be dealt by means of elementary number theory. For instance, maybe we can inductively define the sequence $\{p_n\}$ of (non necessarily prime) numbers such that if (for sufficiently large $n$) each of $p_1,\dots, p_n$ divides $N$ then there exists $p_{n+1}$ coprime with each $p_i$, $i\le n$ such that for each non-zero residue $N’$ modulo $p_{n+1}$ there exists a subproduct $d=p_{i_1}\dots p_{i_k}$ such that $N/d+d>p_{n+1}$ (resp. $|N/d-d|>p_{n+1}$) and $N’/d+d|N$ (resp. $(N’/d-d)|N$). $\endgroup$ – Alex Ravsky Apr 3 '15 at 11:47

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