Problems & Puzzles: Conjectures

Conjecture 52 : SMR Conjecture about consecutive primes

Sebastián Martín Ruiz has found empirically the following simple relationship between consecutive primes p, q:

Floor[(p^2+q^2)/(2*q)]=p

Later he has proved that this relationship follows from another well known conjecture about consecutive primes.

Question.

Can you obtain the SMR relationship from another conjecture or theorem about consecutive primes?

Contributions came from: Patrick Capelle, Farideh Firoozbkaht, Luis Rodríguez, Werner Sand, Andreas Höglund.

***

Patrick wrote:

I prefer the following formulation :
floor(p/q + q/p) = 2.

Proof :
By Chebyshev's theorem there is always a prime between n and 2n for n > 1.
Hence, p < q < 2p when p and q are consecutive primes.
So, p/q < 1 < q/p < 2, which implies that p/q + q/p < 3 
On the other hand, (p - q)^2 > 0, which implies that p^2 + q^2 > 2pq.
Hence, p/q + q/p = (p^2 + q^2)/(p.q) > (2p.q)/(p.q). So, p/q + q/p > 2 
Finally, by  and , 2 < p/q + q/p < 3. Hence, floor(p/q + q/p) = 2.

Some remarks :
1. floor(p/q + q/p) = 2
<==> floor((p^2 + q^2)/pq) = floor(2)
<==> floor((p^2 + q^2)/pq * p/2) = floor(2 * p/2)
<==> floor((p^2 + q^2)/2q) = floor(p)
<==> floor((p^2 + q^2)/2q) = p (SMR formulation).
2. For p < q, floor(p/q) = 0, floor(q/p) = 1 and floor(p/q + q/p) = 2.
3. lim(p/q + q/p) = lim(p/q) + lim(q/p) = 1 + 1 = 2, as p and q approach
infinity.

***

Farideh wrote:

Remi Eismann'conjecture(see A124129):

All prime p for which there are no primes between p
and p+sqrt(p) are: 3, 7, 13, 23, 31 & 113.

So according to this conjecture if p & q are two consecutive
primes and p doesn't belong to A = {3, 7, 13, 23, 31, 113} then
q<p+sqrt(p)<p+sqrt(2q).
Also if p is in A we have q<p+sqrt(2q).
Hence if this conjecture is true then for all primes p we have
q<p+sqrt(2q) or (q-p)^2<2q.
So (p^2+q^2-2p*q)/(2*q)<1 or (p^2+q^2)/(2*q)<p+1 also since
(p-q)^2>0 we have p<(p^2+q^2)/(2*q) hence p<(p^2+q^2)/(2*q)<p+1.
From the last inequality we deduce that  Floor[(p^2+q^2)/2q]=p.

***

Luis wrote:

I found that the SMR conjecture depends on Cramer's.
So: q^2 + p^2 = (q - p)^2 + 2pq

(q^2 + p^2)/2q = (q -p)^2/2q + p

(q - p) = Difference between consec. primes = d. Naturally, q > p

If I can demonstrate that d^2 / 2q < 1 then the SMR conjecture is proved. Because floor[e + p] = p if e < 1.
Cramer says: d / (log(p)^2 < 1
By greater reason d^2 /(log(q))^4 < 1
But 2q > (log(q))^4 for q > 1349

Testing all the differences d between consecutive primes,d from 2 to 1349, we find that d^2 / 2q < 1
Then the conjecture was proved on Cramer's.

***

Werner wrote:

Floor is only a trick to avoid the limes. I am working with the limes:

From PNT we know that the n-th prime number is asymptotically p(n) = n*ln n . Let be p(n)=p and p(n+1)=q two consecutive prime numbers, thus p ~ n*ln n und q ~ (n+1)*ln(n+1).

Then q/p ~ (n+1)*ln(n+1) / (n*ln n) = (n*ln(n+1)+ln(n+1)) / (n*ln n) = (ln(n+1)) / ln n + (1/n)*(ln(n+1)) / ln n.

ln(n+1) / ln n = (ln n + ln(1+1/n)) / ln n = 1 + (ln(1+1/n)) / ln n. Hence

lim((ln(n+1)) / ln n), n->inf = 1. Hence

lim (q/p), n->inf  = lim (1+1/n), n->inf = 1 and

lim (p/q), n->inf = 1

qed

Now once more the proof of Conjecture 52:

lim (1/2*p/q + 1/2*q/p) = 1

lim ((p² + q²)/2pq) = 1

lim ((p² + q²)/2q) = p

qed

***

Andreas wrote:

Mensaje: If we write 2 consecutive primes as p and p+n, where n is the prime gap:

Floor[(p^2+(p+n)^2]/(2*p)] = p + n =>
Floor[ p + n + n^2/(2*p) ] = p + n =>
n^2/(2*p) < 1 => p > n^2/2

According to http://mathworld.wolfram.com/PrimeGaps.html:
Cramér (1937) and Shanks (1964) conjectured that p(n) is approximately exp(sqrt(n)), and Wolf that p(n) is approximately sqrt(n)*exp(sqrt(n)). Both are bigger than n^2/2 for all primes p>=2.

SMR and Wolf conjecture fails for 7 and 11 though, but they show the asymptotic behaviour of p(n).

***

Farideh found that the Capelle's proof and the Sand's proof contain an invalid step:

...both of these two proofs aren't acceptable:

1. The line " floor((p^2 + q^2)/pq) = floor(2) <==>
floor((p^2 + q^2)/pq * p/2)= floor(2 * p/2) " of the first proof isn't acceptable
and it needs to be proved. Because in general case if  floor(x) = floor(y) we
can't deduce that floor(x*z) = floor(y*z) , for example x=7/3, y=2 & z= 7/2.

2. The line " lim((p^2+q^2)/(2pq)) = 1 ==> lim((p^2+q^2)/(2q)) = p " in the
second proof isn't correct, because lim((p^2+q^2)/(2q)) = infinity = lim(p).
From lim((p^2+q^2)/2pq) = 1 and the fact (p(n+1)-p(n))^2>0, we can
only say that for each epsilon>0 there exist a number m such that if n>m
then 1 < (p(n)^2+p(n+1)^2)/(2p(n)p(n+1)) < 1+epsilon.
So, p(n) < (p(n)^2+p(n+1)^2)/(2p(n+1)) < p(n)+epsilon*p(n)  and from this
inequality we can't deduce that floor(p(n)^2+p(n+1)^2)/(2p(n+1)) = p(n).

From these two proofs we only find that floor((p^2+q^2)/(pq)) = 2.

It's easy to show that if a*floor(x) is an integer and  0<a<1 then
a*floor(x) = floor(a*x). Now take a = 1/2 and x = (p^2+q^2)/(pq)  where p & q are two arbitrary consecutive primes(p<q) then we deduce that floor((p^2+q^2)/(2pq)) = 1.

Hence the SMR Conjecture is equivalent to the following nice relation.

p * floor((p^2+q^2)/(2pq)) = floor(p * (p^2+q^2)/(2pq))

Also we have the similar conjecture:
floor((p^2+q^2)/(2p)) = q  where p<>7 and p & q are two consecutive
primes(p<q). Similarly this conjecture is a result of Remi Eismann's conjecture and is
equivalent to the following nice relation.

q * floor((p^2+q^2)/(2pq)) = floor(q * (p^2+q^2)/(2pq))

***

Werner points out that the conjecture 52 applies not only to neighbouring
prime numbers, but to all positive real numbers a,b with b-a < 1+sqrt
(2a+1). The proof for this results simply from the condition. The statement
for prime numbers remains a conjecture. It can now be formulated more
clearly this way:

p(n+1) - p(n) < 1 + sqrt(2p(n) + 1).

***

Maximilian Hasler wrote (May 2007):

We have : floor( (p^2+q^2) / 2q ) =p
<=> floor( (p^2+q^2) / 2q -p ) =0
<=> (p^2 - 2pq +q^2) / 2q < 1
<=> (q-p)^2 < 2q
<=> q-p < sqrt( 2q )

The conjecture should be written in that way, which is obviously the
simplest and clearest way of writing it. As shown, this is a result from Cramer's conjecture ( q-p < log(p)^2 ) or from the conjecture q < p + sqrt(p) for p>113.

Conversely, it implies Andrica's conjecture
http://en.wikipedia.org/wiki/Andrica%27s_conjecture
(thanks to Patrick Capelle for correcting an error):
sqrt(q) - sqrt(p) < 1 (Andrica's conjecture)
<=> sqrt(q) < 1 + sqrt(p)
<=> q < 1 + p + 2sqrt(p)
<=> (q-p-1)^2 < 4p
<=> q^2 + p^2 - 2pq - 2q + 2p +1 < 4p
<=> (q-p)^2 < 2q + 2p - 1
Thus we can see that SMR's conjecture implies Andrica's conjecture.

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