Problems & Puzzles: Conjectures

Conjecture 46. P-Q=R-S

Javier Falcó Benavent sent the following conjecture:
 
If 2 < q < p are prime numbers, then there are two other prime numbers r and s with p < r < 2p  and  
p-q=r-s

Question: Prove it or show it false.

 

Contributions came from Patrick Capelle.

Patrick wrote:

Counterexamples :
p = 5, q = 3
p = 7, q = 3
In the both cases s = p.
I think it would be preferable to change the conditions on p (e.g. p > 7, if there are not other counterexamples)  or on s  (i.e., no restriction for s, which means that s can be equal to p).

This conjecture can be seen with the eyes of Goldbach : even numbers expressed as sums of two primes in two different ways, p+s and r+q, but with particular conditions on the prime numbers to find out. Note that all the even numbers are concerned here because p and q can be arbitrarily selected ...

It was already conjectured that every even number is the difference of two prime numbers in an infinity of ways.
In this context it will undoubtedly not be a surprise if i propose the following generalization for the Benavent's idea :
Let p, q be odd prime numbers.
Then there are an infinity of prime numbers ri and si , with i natural number, such that
p <  r1 <  2.p  <  r2  <  4.p  <  r3  <  8.p  < ... <  ri  <  2i.p  < ... 
and   p - q = r1 - s1 = r2 - s2 = ... = ri - si = ...

Later he added:

The approach with the sums gives an interesting result, which leads to a confirmation of the Falcó Benavent's conjecture (modified in the sense that s can be equal to p) :
All positive even integers > 8 (but different of 12) can be expressed as the sum of two primes in two different ways such that the highest prime number of one sum is greater than the highest prime number of the other sum and smaller than its double.  [1]
If n = 5 or n > 6, then 2n = p+s = r+q  and  p < r < 2p , with p, q, r and s odd prime numbers and  p >= s, r > q. 
 
r is never equal to q when s < p < r.
r is never equal to q  when s = p, because two different sums are concerned here. 
r is never equal to p for the same reason (in the context with p >= s and r > q).
Hence the highest prime number of one sum will be greater than the highest prime of the other sum :  p < r  or  r < p.
In both cases we will have that r < 2p :
p >= s  ==>  p >= (2n)/2 = n  ==>  2p >= 2n.   
q different of 0 ==> r < 2n  ==>  r < 2p.
Note that we have also p < 2r, because  r > q  ==>  r > (2n)/2 = n  ==> 2r > 2n  and  p < 2n  ==>  p < 2r.
 
Example :
18 = 11 + 7 = 13 + 5. 
As you can see p = 11, r = 13  or  p = 13, r = 11, giving  p < r  or  r < p (so, by convention we can choose p < r  for the presentation of the situation), but  in both cases we have  r < 2p  and  p < 2r.
 
So, the fact that 2n = p+s = r+q , with p >= s  and  r > q,  implies that  r < 2p  and  p < 2r.
It's also true when an even integer is expressed as a sum of two primes in more than two different ways in so far as the comparison between prime numbers involves any two sums.
 
In conclusion, the proposal [1] is not different or more useful than the simple conjecture saying that 'all positive even integers > 8 (but different of 12) can be expressed as the sum of two primes in two different ways', because the second part of the proposal [1] is a consequence of the first part. The way of obtention of the prime numbers in the Falcó Benavent's conjecture and the transformation of the differences into sums (or reverse) do not change this conclusion basically : the property " p < r < 2p " appears automatically when we express an even integer as sum of two primes in two different ways.

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Antoine Verroken wrote (Set., 2006)

Please will find below a new attempt for c.46 :

1. p – q = r – s           2 < q < p             p < r < 2p
2. p – q = d       d is even 

a.       d = 2

because of the constellation  3,5,7 and the fact that the constellation of
q,q+2,q+4 can only exist for q = 3,the conjecture can only be correct for
q>3.For q>3 we have to prove that the number of twin primes between
q and 2p is larger than or equal to 2.According to the conjecture of Hardy
and Littlewood there are minimum 2 twin primes between q and 2p.

            b.    d > 2n   ( n = 1 .. infinity )

there are several types of triplets f.i. 17,23,29 for d=6.Thus we need mini
mum 3 primes escorted at a distance of d by another prime. The number
Nd of such primes is given by the conjecture of Polya : 

Nd = N2 x ( Product of ( p – 1) / ( p – 2 )) for all p :  odd primes which divide d.
N2  :number of twin primes. Product is always larger than 1 (  d(10)  Prod. = 1.33 )

The number of twin primes between ( p – 2 ) and 2p is larger than or equal to 3 for p > 29.

For q=5 to p=29 we can see by computation that the conjecture sounds for d > 2n.

Conclusion : except for q = 3 and d = 2 the conjecture can be proved if the conjectures
of Hardy & Littlewood and Polya sound.

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