Gold bach Reloaded

Inferences from Strong Gold Bach’s Conjecture

Strong Gold Bach’s Conjecture is as given below:
e  even number
p  prime number
o  odd number
According to Strong Gold Bach Conjecture,
Any even number >=8 can be written as a sum of two different primes.

Now let’s get into my observations...
Using Strong Gold Bach’s conjecture
Any e = p1+p2, where p1 and p2 are two different prime numbers

Now take any prime number p
2p is an even number,
So 2p = p1 + p2 using Strong Goldbach’s conjecture where p1 and p2 are two different primes

So,

p = (p1+p2)/2

Inference1:

Therefore any prime number can be written as the average of two different primes.

E.g.:
Take 17
2*17 = 34 = 31+3
Therefore 17 = (31+3)/2
This also means that any prime number is equidistant from two some other two primes
This inference shows the uniformity of the primes though they can’t be represented by a single formula

Inference2:

Since p = (p1+p2)/2, using the above inference,
p1 can be a maximum of 2p-1 so that the other prime can be a minimum of 1

From this we can see that if p is any prime, there is at least one prime number between p and 2p

E.g.:
Take p =
31 and 2*31 = 62
Between 31, there are primes 37, 43 etc...

Inference3:

According to Strong Goldbach’s conjecture, any even number
e = p1 + p2, where p1 and p2 are two different prime numbers
But,
e = o1 + o2, where o1 and o2 are two non prime odd numbers
So basically, o1 + o2 = p1 + p2
Hence, suppose o1 = p1 + x, where x is some difference
Then o2 =p2 –x, since, in the end the (p1+x) + (p2-x) = p1+p2 = o1+o2
Hence we can see that
For any two pair of odd numbers, o1 and o2, you have two prime numbers p1 and p2, such that
p1Similarly, there is another group, o1
So, any pair of odd numbers has a pair of symmetrically placed primes

E.g.: Take two odd numbers, say 15 and 57. Here the primes below 15 are 3, 5, 7, 11 and 13. So I just take the differences of this number with 15 and add those to 57, one of them will be definitely a prime.
So 57 + (15-3) = 69  not prime, 57 + (15 – 5) = 67  Prime! So the numbers are 5, 15, 57 and 67

Inference4:

There are two primes equidistant from any number!!

According to Strong Goldbach’s conjecture, any even number
e = p1 + p2, where p1 and p2 are two different prime numbers.
So p1 = e/2 –k and p2 =e/2 + k

Hence we can see that this points to the fact that there are two primes equidistant from any number ,as the e/2 can be an odd or even

Eg: Taking a76, we have 73 and 79. Taking 109, we have 79 and 139 and equidistant primes from 109

This is useful if we know all primes lesser than a number N and then we just have to check if the numbers N+(N-pi) are primes, where pis are primes below N