Problems & Puzzles:
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
Question: Prove it or show it false.
Contributions came from Patrick Capelle.
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
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
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 =
s2 = ... =
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. 
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.
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
In conclusion, the proposal  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  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.
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
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
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
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.