A New Conjecture

[AntonVrba]

Hi,

Earlier this year when the new largest prime was announced, it re-stimulated my interest in number theory and prime numbers.

In investigating primes in the quest to understand them (silly me), I discovered a new property (that means I have not come across it before and it might be known any way).

Let N be an even number and Q an odd number such that
P1 = Abs(N-Q) and P2 = N+Q are both prime

Conjecture 1:
For every even number N there are infinite pairs of primes P1, P2
such that either the sum of P1 and P2 divided by two equals N,
or the difference of P1 and P2 divided by two equals N .

Conjecture 2: GCD(N,Q) = 1

Conjecture 3:
The minimum number of distinct factors of N and Q is three.

That means if Q = p^y then N has at minimum factors (2 and x)
and if N=2^y then Q has at minimum factors (3 and x)


Conjecture 4:
If N=p# then the smallest value of Q is Prime.
or
If N=n! then the smallest value of Q is Prime.

Conjecture 5:
Iff Px and Py are base 2 probable primes,
with Qx =(Px-Py)/2 and Nx = (Px+Py)/2
and Qx > largest factor of Nx
and Qx is the minimum solution to finding Px and Py,
and GCD(Qx, Nx)= 1 ,
then both Px and Py are prime.

I have named this the Even-Prime-Balance Conjecture and defined the EPB function

EPB(N, Q)= True , Iff N+Q is Prime AND N-Q is Prime

You can look at the conjecture at (basically the same as above)
http://www.geocities.com/al_at_i/epb.html

Your comments would be much appreciated (anton@a-l-v.net) and has the EPB conjecture any significance in basic number theory

The conjecture is based on research of small and large N.
I have tested above for many N, and checked it using Pa = pseudo primes and Pb being the nearest prime (larger and smaller) and found above to hold.

best regards
Anton

[R.D. Silverman]

Originally Posted by AntonVrba

Hi,

Earlier this year when the new largest prime was announced, it re-stimulated my interest in number theory and prime numbers.

In investigating primes in the quest to understand them (silly me), I discovered a new property (that means I have not come across it before and it might be known any way).

Let N be an even number and Q an odd number such that
P1 = Abs(N-Q) and P2 = N+Q are both prime

Conjecture 1:
For every even number N there are infinite pairs of primes P1, P2
such that either the sum of P1 and P2 divided by two equals N,
or the difference of P1 and P2 divided by two equals N .

Conjecture 2: GCD(N,Q) = 1

Conjecture 3:
The minimum number of distinct factors of N and Q is three.

That means if Q = p^y then N has at minimum factors (2 and x)
and if N=2^y then Q has at minimum factors (3 and x)


Conjecture 4:
If N=p# then the smallest value of Q is Prime.
or
If N=n! then the smallest value of Q is Prime.

Conjecture 5:
Iff Px and Py are base 2 probable primes,
with Qx =(Px-Py)/2 and Nx = (Px+Py)/2
and Qx > largest factor of Nx
and Qx is the minimum solution to finding Px and Py,
and GCD(Qx, Nx)= 1 ,
then both Px and Py are prime.

I have named this the Even-Prime-Balance Conjecture and defined the EPB function

EPB(N, Q)= True , Iff N+Q is Prime AND N-Q is Prime

You can look at the conjecture at (basically the same as above)
http://www.geocities.com/al_at_i/epb.html

Your comments would be much appreciated (anton@a-l-v.net) and has the EPB conjecture any significance in basic number theory

The conjecture is based on research of small and large N.
I have tested above for many N, and checked it using Pa = pseudo primes and Pb being the nearest prime (larger and smaller) and found above to hold.

best regards
Anton

Conjectures 1 and 2 are trivially true. (2 line proofs each) 3 follows
immediately from 2.

4 is certainly false. Put N = 101#. It is virtually certain
that there exists p1, p2 , each > 101, such that N+p1p2 and N-p1p2
are both prime.


I do not understand what you are trying to say in 5. Please clarify.

[AntonVrba]

Originally Posted by R.D. Silverman

4 is certainly false. Put N = 101#. It is virtually certain
that there exists p1, p2 , each > 101, such that N+p1p2 and N-p1p2 are both prime.

101# + Q and 101# - Q are prime for
Q=523, 739, 1307, 2971, 3709, 3889, 5981, 7393, 7879, 10909, 12757, 14369, 107x139, 16333, 16831,18523, 19273, 19979, 21017, 21937, 23131, 24251, 27751, 107x271, 29759 etc etc

The smallest value of Q is prime hence conjecture 4 is true, the conjecture does not say each Q is prime

Originally Posted by R.D. Silverman

I do not understand what you are trying to say in 5. Please clarify.

Check Fermat's Little Theory, Carmichael numbers and pseudoprimes. Conjecture 5 would simplify prooving numbers as prime for the cases where C5 can be demonstrated.

[AntonVrba]

Originally Posted by R.D. Silverman

I do not understand what you are trying to say in 5. Please clarify.

I have re-read conjecture 5 and like to add additional clarrifications as I possibly have not expressed my self clearly enough

Conjecture 5:
Iff Px and Py are base 2 probable primes,
with Qx =(Px-Py)/2 and Nx = (Px+Py)/2
and Qx > largest factor of Nx.

that should be clear, Px and Py are two odd numbers shown as probable prime satisfying Fermats Little Theorem.

and Qx is the minimum solution to finding Px and Py,

there are no smaller Qx such that Nx+Qx and Nx-Qx are both prime, (This is better wording but this contstaint might not be necesarry)

and GCD(Qx, Nx)= 1 ,
then both Px and Py are prime.

and this could be a new prooving method to declare Px and Py as Prime.

[R.D. Silverman]

Originally Posted by AntonVrba

101# + Q and 101# - Q are prime for
Q=523, 739, 1307, 2971, 3709, 3889, 5981, 7393, 7879, 10909, 12757, 14369, 107x139, 16333, 16831,18523, 19273, 19979, 21017, 21937, 23131, 24251, 27751, 107x271, 29759 etc etc

The smallest value of Q is prime hence conjecture 4 is true, the conjecture does not say each Q is prime



Check Fermat's Little Theory, Carmichael numbers and pseudoprimes. Conjecture 5 would simplify prooving numbers as prime for the cases where C5 can be demonstrated.

Conj 4 is still likely to be false on probabilistic grounds. I am sure there
is some p such that the smallest Q will be composite. We just haven't
look high enough.

And I still don't follow what you are trying to say in Conj 5. Please
expand upon the conjecture.

[alpertron]

A proof of conjecture 4 is very difficult to obtain.

But if some conjectures about prime gaps are valid, for instance G(n) < (ln n)^2 (see http://mathworld.wolfram.com/PrimeGaps.html), then the conjecture 4 is valid also.

This is because ln(n#) < n, so the gap is less than n^2. The difference Q between the first prime after n# and n# must be less than n^2.

Notice that Q does not have a divisor m<n because n# is multiple of m also, in which case n#+Q would be also multiple of m, so it would not be a prime.

Thus Q must be prime.

[R.D. Silverman]

Originally Posted by alpertron

A proof of conjecture 4 is very difficult to obtain.

But if some conjectures about prime gaps are valid, for instance G(n) < (ln n)^2 (see http://mathworld.wolfram.com/PrimeGaps.html), then the conjecture 4 is valid also.

This is because ln(n#) < n, so the gap is less than n^2. The difference Q between the first prime after n# and n# must be less than n^2.

Notice that Q does not have a divisor m<n because n# is multiple of m also, in which case n#+Q would be also multiple of m, so it would not be a prime.

Thus Q must be prime.

Put N = p#

There is a requirement that N+Q *and* N-Q both be prime.
Certainly (by Cramer's conj as you suggest), the first prime gap after N must
be less than p^2. Thus Q can't be p1*p2 with p1, p2 > p if we just consider N+Q. But will N-Q also be prime?

[AntonVrba]

Originally Posted by R.D. Silverman

Put N = p#

There is a requirement that N+Q *and* N-Q both be prime.
Certainly (by Cramer's conj as you suggest), the first prime gap after N must
be less than p^2. Thus Q can't be p1*p2 with p1, p2 > p if we just consider N+Q. But will N-Q also be prime?

Using the previous example
101#+k is prime for
k=-131, -139, -149, -167, +233, -239, -461, -463, -491, +523, -523 (the negative bias is just coincidental

BTW
173#+-191 are both Prime and no other primes between these two, whereas
293#+-10987 are both prime with 93 primes between these two and all for all 293#+k the k's are prime (for k's negative or positive)

[alpertron]

Bob, you are right. It appears that at 11 PM I only write gibberish^TM and illucid^TM statements.

If both n#-Q and n#+Q are primes, Q can be possibly greater than (ln n)^2, invalidating my previous argument.

So conjecture 4 is stronger than Cramer's conjecture. Notice that still nobody found a prime gap near to (ln n)^2 so it is possible that conjecture 4 is true too.

[AntonVrba]

You got me thinking

Here is conjecture 6:

Originally Posted by AntonVrba

For Primes p#+Q or p#-Q and Q<p^2 then Q is prime

Here is conjecture 7:

Originally Posted by AntonVrba

For Primes p#+Q or p#-Q and Q composite, the factors of Q are larger than p (except for 3# and 5#)

Here are results for the first Q that is composite for p#+-Q prime
97#+107x109
101#-103x109
103#+107^2
107#-109^2
109#+113x191
113#+127x131
127#-131x139
131#+149^2
137#-139x151
139#-149x151
149#-157x163
151#-163x167
157#-163x181
163#-173x181
167#-179x193
173#-193x199
179#-181^2

[R.D. Silverman]

Originally Posted by alpertron

Bob, you are right. It appears that at 11 PM I only write gibberish^TM and illucid^TM statements.

If both n#-Q and n#+Q are primes, Q can be possibly greater than (ln n)^2, invalidating my previous argument.

So conjecture 4 is stronger than Cramer's conjecture. Notice that still nobody found a prime gap near to (ln n)^2 so it is possible that conjecture 4 is true too.

On the contrary. Your writing was concise, clear, and cogent. You just
missed a condition.

Here's what makes me think the conjecture is probably false.

Put N = p#. The probability that N+Q is prime is about 1/log(N+Q) ~ 1/p
The probability that N-Q is also prime is ~1/p. We want to search over
values of Q, so that both are prime. If we let Q go from 1 to K,
then the probability of finding both prime for some Q is

sum from Q = 1 to k of 1/p^2 and this is just k/p^2 which is small.
for k ~ log^2 N. I expect that for some p's we will have to take k to
be bigger than log^2 N, i.e. Q will be p1*p2 for p1,p2 > p.

The problem is that 1/p^2 is quite small for large p.