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If $N=2.m^n+1$ (where $m$ is prime) you can prove if $N$ is prime or not by these two steps:

Step (1) if $a^{2.m^{n-1}}=L \mod(N)$ (which is $L\neq1$ )

Step (2) $L^{m}=1 \mod(N)$
So N is prime.

The proof.

If (N) is composite so all of his factors are smaller than (N), so to achieve step (1) it requires that (m^n) divides the (x) for one of (N) factors, which is impossible, because of that the smallest number that could be prime and (m^n) divides (x) for it , is (2.m^n+1).

Finally : I know that you don't know what do I refer to with (x), ok (x) is the smallest value that achieves that (a^x=1 mod(p)) , which is be different from base to another, for example:

x(13) = 4 when the base is (5) because 5^4=1 mod 13

Another example.
x(13)=12 when the base is 2 , because 2^12=1 mod 13.

These are some numbers I have proved that they are primes using this test :

2.3+1 the used base is 2

2.3^2+1 the used base is 2

2.3^4+1 the used base is 2

2.3^5+1 the used base is 2

2.3^6+1 the used base is 3

2.3^9+1 the used base is 3

2.3^16+1 the used base is 2

And I wanted to say that when I applied my test on this number for example :

2.3^4+1=163 there were a 108 bases that can proof that this number is prime

But when I applied lucas test on this number there were only 54 bases that can proof that this number is prime

So which one is better for this form ? And how can I publish this test ?

I forgot to say that this algorithm can be generalized when :

N=ABCD....Z.m^n+1 when :

(A,B,C.....Z and m are primes, and m is larger than ABCD....Z)

for example : 727=2.3.11^2+1

But it is not generalized for the same reason

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closed as unclear what you're asking by Théophile, N. S., Peter, Shaun, John Omielan May 16 at 7:23

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