Symmetry method to find Primes.

By observation I found this method to find primes which can be proofed easily.

THEOREM

For every $p_n$ & $a$ that satisfy $p_n < p_n\#-a < p_{n+1}^2$ where $(a > p_n)$

If $(a)$ is a prime then $(p_n\#-a)$ is also a prime

and if $(p_n\#-a)$ is a prime then $(a)$ is possible a prime

Example:
$a=2213$ (prime number)

let $p_n$ any prime satisfy the above

for instance $p_5=11$ then $p_5\#=2310$

$11<2310-2213<13*13$

$11<97<169$ it's OK

Then 97 must be a prime.

THE PROOF

$(a)$ and $(p_n\#-a)$ are symmetrical around $\frac {p_n\#}{2}$

Let $a=\frac {p_n\#}{2}+b$ then $(p_n\#-a)=\frac {p_n\#}{2}-b$

$p_n\#=(\frac {p_n\#}{2}+b)+(\frac {p_n\#}{2}-b)$

assume that $(\frac {p_n\#}{2}+b)$ is divisible by $\{p_k : p_k \le p_n\}$

then $(\frac {p_n\#}{2}-b)$ must be divisible by $p_k$ too

because $\frac {p_n\#}{2}$ is always divisible by any prime $2 < p_k \le p_n$ so if $(+b)$ is divisible by $p_k$ then $(-b)$ is divisible as well. and if $(\frac {p_n\#}{2}+b)$ is not divisible by any prime of $\{p_k : p_k \le p_n\}$ then $(\frac {p_n\#}{2}-b)$ should be also not divisible by the same prime. So we have this range which we can define the symmetry of the numbers that are not divisible by $\{p_k : p_k \le p_n\}$. that range is from $(p_n+1)$ to $(p_n\#-(p_n+1))$ then $a$ should always greater than $p_n$ $(a > p_n)$

but unfortunately this range of symmetry is broken by the primes greater than $p_n$ because $p_n\#$ is not divisible by $p_{n+1}$ and greater.

So the limits of symmetry ,which can not be affected by the primes greater than $p_n$ , can be defined between $p_n$ and $p_{n+1}^2$

So if $(p_n\#-a)$ satisfy $p_n \lt p_n\#-a \lt p_{n+1}^2$ then it is in the range of symmetry and if $(a)$ is prime then $(p_n\#-a)$ is also prime

but if $(p_n\#-a)$ is prime then $(a)$ could be prime because it will not be divisible by $\{ p_k : p_k \le p_n\}$ but on the other hand it could be divisible by the primes greater than $p_n$ because it is may located out of the range where the symmetry is broken.


We can notice the symmetry in these extra examples

1) For $p_3\#=2*3*5=30$
$p_{n+1}^2=7∗7=49$
so all the primes $5 < p < 30$ are symmetrical around 15 $\frac {p_n\#}{2}=15$ note that $15±2 , 15±4 , 15±8$ are all primes.

2) For $p_4=210$ $p_{n+1}^2=11∗11=121$, number $121$ is less than $210$ which will break the symmetry starting from $121$ so the range that will maintain the symmetry is $(7 < 210-a < 121)$ , $a > 7$
$\frac {p_n\#}{2}=105$ note that $105±2 , 105±4 , 105±8$ are primes etc...
Let $a=151$ (prime number) then $210-151=59$
$7 < 59 < 121$ then $59$ is a prime.