Conjecture 52 : SMR Conjecture about consecutive primes

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

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.

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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 [1]
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 [2]
Finally, by [1] and [2], 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.

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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.

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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.

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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

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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).

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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))

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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).
 

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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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