Werner D. Sand sends the following conjecture from his own
invention:

Let A, B be positive integers such
that:

a) A, B are coprimes

b) A*B contains at least once each prime factor from 2 to Q

Then P=A+B or P=abs(A-B) is prime if 1<P<R², where R is the next larger

prime to Q. (This can be proven).

**CONJECTURE: Every prime number
P>3 can be produced in **

this way.

Question: Prove it or show it false.

Luke
Pebody proved this conjecture.

***

Luke wrote:

Let P be a prime greater than 3. Let
Q be the prime immediately before P.

Let B=2*3*5*7*...*Q, A=P+B. Clearly A,B are coprime and AB contains
exactly once each prime factor up to Q and P=Abs(A-B).

***

Mr. Sand's original conjecture
implied that "A*B contains at least once each prime factor from 2 to
Q and no factor else.", but in the final statement the “and no
factor else” condition was missed. He thinks that in the original
statement his conjecture is not provable.

***

Daniele Degiorgi wrote:

...It seems that for these
conjectures, like the last ones (at least 47 and

46), there are a large number of possibilities, and thus, like the
Goldbach

conjecture, they are surely true for almost all integers. It seems that
we

do not yet have appropriate tools to prove the absence of exception,
while

finding some, if any, could be a hopeless work.

Note also that if for the Goldbach conjecture a counterexample could be

checked examining a finite number of cases. For this conjecture this is
not

possible and an explicit proof is needed.

Take for example R=5 and all possible A,B. Which primes are not

representables in this way? The first candidate is 103 but I am not able
to

prove that no m,n exists such that abs(2^n-3^m)=103 (it is easy to see
that

no m,n exists with 2^n+3^m=103, or 2^n*3^m+1=103, or 2^n*3^m-1=103).

Does such a proof exist?

***

Regarding the validity of the Luke's
proof, Degiorgi wrote on my request:

It depends on how the original
conjecture is interpreted. I interpreted b) as not allowing divisors
larger than Q. In the solution proposed by Luke A*B has also some
divisors larger than Q. May be that Werner could precise this [Done
above]

...

Without such limitations, the solution of Luke is correct. In fact in
this

case there are infinite solutions as you could multiply B by any power
of

any prime different of P still obtaining the same effect.

***

Werner wrote:

For
P=103 the prime numbers 2,3,5,7 are admitted, because 11 ² =121>103,

thus Q=7. Thus e.g. is 103=3*5*7-2. Only to use 2 and 3 would not be

sufficient, because R=5 and 103>5². Only to use 2,3,5 would not be

sufficient either, because R=7 and 103>7². 103 is definitely prime only
if

all prime factors < sqrt (121) were used. The problem is a
generalization of

the question whether there are further solutions for abs (3^m-2^n) =1

besides 3²-2³=1 (Catalan's conjecture, proved 2002 by Mihailescu). With

increasing P it is becoming more and more difficult to use all prime
factors

< R since the shears go more and more apart e.g. between 3^m and 2^n.
That

it is impossible starting from a certain P, is to prove. Perhaps
Mihailescu's

proof can help.

***