Is this new method to find primes? By observation I found this method which can be proofed easily ,but not sure if it is mentioned somewhere before and that's what almost should be.
This is the method:
For every $p_n$ & $a$ that satisfy $p_n < p_n\#-a < p_{n+1}^2$
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
...
EDIT:
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The Proof
Firstly I miss the condition of $(a)$ which should always greater than $p_n$ $(a > p_n)$ as will be mentioned in the following 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.
My question is:
I need to varify this proof numerically for large numbers and check if it is correct for large prime numbers
We can notice the symmetry in these 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.
 A: Your first test is false: $n=3$ and $a=2$. Then $p_n = 5$, $p_n \# - 2 = 28$ is obviously composite, but $2$ is prime.

For your second test: as you say, $p_n \# - a$ being prime doesn't necessarily mean $a$ is prime: see $n=4, a=121$, where $p_n \# - a = 89$ is prime (and less than $p_5^2 = 121$) but $a = 121 = 11^2$ is not.
So that part doesn't constitute a reliable way of finding primes, and I can't see that it can be easily improved.

I originally wrote the following about Part 1, before finding that counterexample.
For your first test, this is a way of finding primes. I'm going to go by contradiction, I'm afraid, though it's aesthetically displeasing.
Suppose $p_n \# - a$ were composite; then it would be divisible by some prime $p$.
But because $a$ is prime, we must have either $p > p_n$ or $p=a$.
So the only primes that divide $p_n \# - a$ are $a$ or things bigger than $p_n$.
If $p_n \# - a$ were divisible by two primes (counting multiplicity) bigger than $p_n$, then your second condition would fail: $p_n \# - a < p_{n+1}^2$ would be false.
So the prime factorisation of $p_n \# - a$ is "at most one prime bigger than $p_n$, and some number of copies of $a$".


*

*If $a > p_n$, then $a$ can't appear in the prime factorisation at all, because $a \not \mid p_n \#$ so $a \not \mid p_n \# - a$.
That means the prime factorisation is just "some single prime bigger than $p_n$", contradicting compositeness of $p_n \# - a$.


So $a = p_i$ for some $i$, and $$p_n \# - a = p_k^{\epsilon} \times a^r$$ for some $k > n$ and some $r \geq 0$, where $\epsilon$ is either $0$ or $1$.
