**
Conjecture 52 :** **SMR
Conjecture about consecutive prim****es**

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 [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.

***

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.

***