Problems & Puzzles: Conjectures

Conjecture 48. P=A+/-B

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.

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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).

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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.

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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?

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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.

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