Problems & Puzzles:
A property of
prime twins, only?
Sebastián Martín Ruiz sends the following conjecture from his own
Let p, q be consecutive prime numbers, p<q.
Conjecture: p&q are
prime twins iff z is integer.
Question: Prove it or show it false.
Contributions came from Farideh
Firoozbakht, Danielle G. Degiorgi, Mike Oakes & Robin García.
The following statement,
((p(n+1)^2 + p(n)^2)/2-1))^(1/2) is
integer iff p(n+1) - p(n ) = 2 (**)
is a result of conjecture 30 and it can be mentioned
as a conjecture.
Proof : Let f(n) = ((p(n+1)^2 + p(n)^2)/2 -
1))^(1/2) and d(n) = p(n+1) - p(n)
so f(n) = ((p(n)+d(n)/2)^2 + d(n)^2/4 - 1)^(1/2)
If d(n) = 2 then as SMR pointed out ( by using (I) )
we get that f(n) = p(n)+1
so f(n) is an integer. Now suppose that f(n) is an
integer, then we have the following two cases:
1. d(n) ^2/4 - 1 = 0 , namely d(n) = 2 & f(n) = p(n)
2. d(n) ^2/4 - 1 > 0 or d(n) > 2.
By using the relation (*): d(n) < 2(log p(n))^2
which is one of the results of conjecture 30, we show
that the second case
If d(n) ^2/4 - 1 > 0 (d(n) > 2) then f(n) >= (p(n)+d(n)/2)
p(n) + 3/2 <= (1/8)*(d(n) -
but by using (*) we conclude that,
p(n) + 3/2 < (1/8)*(2(log
p(n))^2 - 2)^2
p(n) + 3/2 < (1/2)*(log p(n))^2
But we can easily show that if p is a number greater
than 1229 then
p + 3/2 > (1/2)*(log p)^2 - 1)^2 which contradicts
the last inequality
and for n < pi(1229) = 201 we can directly check that
if f(n) is an integer then
d(n) = 2. Hence using (*) we can deduce the statement
Danielle sent his contribution in
this .pdf file. In the email he added
the following comments:
include a pdf (and the related latex file) with some notes on
conjecture where it is shown, that two consecutive primes giving an
integer z need to have difference 2 or a difference not smaller than
the Riemann Hypothesis implies that the gap between two consecutive
primes is smaller than p^(1/2+eps), the conjecture probably also
follows from it.
the Riemann Hypothesis is not true, we still known that the gap is
smaller than p^0.525, and thus is very unlikely (but not necessarily
impossible) to find a counterexample, in any case very large.
There are other conjectures implying this: from [2,A8], Dorin
Andrica conjecture imply q-p<sqrt(q)+sqrt(p), and as Bertrand
Postulate shows that q-p<2p, it would follow q-p<2sqrt( 2p).
The "only if" is trivial:
If p and q are twin primes, then q = p+2, so z = (p+1).
As for the "if":
The conjecture is true for p = 2, so assume p >= 3.
If we write
q = p+2*d,
z^2 = (p+d)^2 + (d^2-1).
(p+d+1)^2 is the next perfect square after (p+d)^2.
(p+d)^2 + (d^2-1) < (p+d+1)^2 ... (*)
then z^2 is /not/ a perfect square.
The condition (*) is:
(d^2-1) < (p+d+1)^2 - (p+d)^2
d^2-1 < 2*p+2*d+1
(d-1)^2 < 2*p+3
d < (sqrt(2*p+3) + 1)
q < p + 2*(sqrt(2*p+3)+1)
In words: the gap between the prime p and the next prime q
should not exceed 2*(sqrt(2*p+3)+1).
This is easily verified to hold for all primes p <= 10^9.
It is believed that the gap between successive primes (p, q)
is bounded by sqrt(p) for all p, but this has never been
Conclusion: The Conjecture is almost certainly true, but is
equivalent to another conjecture which is very deep and
unlikely to be proved any time soon.
If p and q are consecutive odd prime, q=p+g and g is even->q=p+2t
If t=1, then p and q are twin primes and z=p+1 is an integer.
If t>1 and z is an integer, then 2t^2-1>t^2 and z>p+t. So z=p+t+n.
p=((t-n)^2-(2n^2+1))/2n, q=p+2t, z=p+t+n
We check t values such that p and q=p+2t are primes, and the number of
primes between p and q:
t=4 p=3 q=11 z=8 2 primes
t=6 p=11 q=23 z=18 3 primes
t=12 p=59 q=83 z=72 5 primes
t=24 p=263 q=311 z=288 7 primes
t=30 p=419 q=479 z=450 10 primes
t=66 p=2111 q=2243 z=2178 15 primes
t=78 p=2963 q=3119 z=3042 16 primes
t=102 p=5099 q=5303 z=5202 21 primes
p=((t-2)^2-9)/4=((t+1)/2)((t-5)/2)is always composite.
t=28 p=101 q=157 z=132 10 primes
t=62 p=577 q=701 z=642 19 primes
t=15 p=11 q=41 z=30 7 primes
t=33 p=101 q=167 z=138 12 primes
t=39 p=149 q=227 z=192 13 primes
t=57 p=347 q=461 z=408 19 primes
t=63 p=431 q=557 z=498 18 primes
t=87 p=857 q=1031 z=948 24 primes
t=14 p=3 q=31 z=22 8 primes
t=24 p=31 q=79 z=60 10 primes
t=66 p=367 q=499 z=438 21 primes
t=84 p=619 q=787 z=708 23 primes
And so on...When n increases, q-p=f(p) increases.
We clearly see that if t>1 and z is an integer, then p and q are not
consecutive primes, because q-p is always of the order of sqrt(p) which
is much more than the average gap between consecutive primes: ln(p)
And so the conjecture is VERY PROBABLY true.
To prove it would be much harder.
Thomas R. Nicely wrote (July 06):
I restate the conjecture as follows.
CONJECTURE: Suppose p and q are consecutive primes, p < q. Define
J = (p*p + q*q)/2 - 1. Then J is a perfect square if and only if
p and q constitute a twin-prime pair.
Proof of the _if_ part is straightforward. For then q=p+2 and
2J = p*p + (p+2)*(p+2) - 2 = 2*p*p + 4*p + 4 - 2, so
J = p*p + 2*p + 1, thus J is the square of (p+1).
I have been unable to prove the _only if_ part. I have verified
it numerically for all consecutive primes < 2^32=4294967296. If
the primes are not consecutive, the assertion is no longer true,
as is shown shown by the counterexamples (p=3, q=11) and
(p=67, q=479), among others.
A sufficient condition for the proof of the _only if_ part appears
to be d < 2*sqrt(p) + 1, where d=q-p=2r (d > r > 0) is the
difference (gap) between consecutive primes and r is half the
difference (we can assume p and q odd, since the conjecture is
obviously true for p=2 and q=3). The logic is as follows:
Suppose d < 2*sqrt(p) + 1 ; (**)
===> (d - 1)^2 < 4p
===> 4p > (d - 1)^2
===> 4p > (d - 1)^2 - 6
===> 4p > (2r - 1)^2 - 6
===> 4p > 4r^2 - 4r - 5
===> 4r^2 < 5 + 4p + 4r
===> 4r^2 - 4 < 1 + 4p + 4r
===> r^2 - 1 < 1/4 + p + r
===> (p + r)^2 + r^2 - 1 < (p + r)^2 + 1/4 + (p + r)
===> (p + r)^2 + (r^2 - 1) < (p + r + 1/2)^2
===> (p + r) <= sqrt((p+r)^2 + (r^2 - 1) < (p + r + 1/2)
The LHS of the last inequality is true because r >= 1;
equality holds if and only if r=1. Thus the nearest integer
to the sqrt must be p+r, so that (p+r)^2 + (r^2 - 1) is a
perfect square if and only if r=1 (in which case d=2 and p
and q are twin primes). Now note that
J = (p^2 + (p+2r)^2)/2 - 1 = (2p^2 + 4pr + 4r^2)/2 - 1
= p^2 + 2pr + 2r^2 - 1 = (p^2 + 2pr + r^2) + (r^2 - 1)
= (p + r)^2 + (r^2 - 1)
Thus J is a perfect square if and only if r=1, d=2, and p and
q are twin primes---IF condition (**) above is met, that is,
d < 2*sqrt(p) + 1 for all consecutive primes p and q. This
amounts to the assumption that the difference between consecutive
primes remains "small" in a sense, or that prime gaps remain
"small" compared to the bounding primes. The "smallness"
requirement is the reason the conjecture fails for some
non-consecutive primes such as (p=3,q=11) or (p=67,q=479).
Condition (**) is known to be true for all consecutive primes
which have been directly calculated---at the moment, for all
p < 4.1*10^17. In fact, for large values of p, d is much
less than sqrt(p); the largest d below 4.1*10^17 is only 1356,
for p=401429925999153707 (Knuth, 2006). Indeed, there exist
much stronger conjectures regarding the asymptotic behavior
of d. Piltz (1884) conjectured that d=O(p^e) for any fixed e > 0;
Cramér (1936-37) conjectured d=O((ln(p))^2). However, as of 1995,
the best _proven_ result was that d=O(p^0.535...) (Baker and
Harman, 1995), and even assuming the Riemann Hypothesis, the
best result proven is d=O(sqrt(p)*ln(p)) (Cramér, 1936-37). Thus
condition (**) remains, for the moment, a conjecture, and
therefore so does the _only if_ part of your own conjecture.
However, (**) is now known to be true for p < 4.1*10^17, and
thus the _only if_ statement is also known to be true throughout
Of course, it may be possible to prove the _only if_ portion
by some other line of reasoning which is not apparent to me.
.. a document an important source of
information regarding the asymptotic behavior of d, the
difference between consecutive primes; namely, "The new book
of prime number records," Paulo Ribenboim (3rd edition, 1996,
Springer-Verlag, New York), pp. 250-265. A more recent edition
exists, but I do not have access to it. Also, the information
on the state of the search for prime gaps is posted at