# Semiprime Antics

hv at crypt.org hv at crypt.org
Sun Oct 12 18:32:38 CEST 2003

```"Chuck Seggelin" <barkeep at plastereddragon.com> wrote:
:Hi there,
:
:I was working on some new sequences last night involving semiprimes and I
:encountered this curiousity.  Take a semiprime N (a number which is the
:product of two primes, see A001358) and its two prime factors J and K and
:test the following rules:
:
:    r1: N+J-K is prime
:    r2: N-J+K is prime
:    r3: N+J+K is prime
:    r4: N-J-K is prime
:
:What I noticed was that for any combination of three (or less) of these
:rules, there are abundant semiprimes that pass the test.  But if one tries
:to enforce all four rules, there appear to be only two: 10 and 15.
[...]
:Is there an obvious reason why no semprimes other than 10 and 15 can adhere
:to all four rules above?

Yes: looking at these equations modulo 3, at least one of them is
divisible by 3 for each combination of J and K unless one of J or K
is itself 3.

Looking again modulo 5 with J = 3, we find that at least one of the
combinations is divisible by 5 unless K = 5.

So the only possible solutions are a) J=3, K=5, and b) small solutions
where the result divisible by 3 or 5 is the prime itself. 10 is then of
the second sort (giving r4 = 3).

Hugo van der Sanden

```