20180204, 20:09  #1 
"Sam"
Nov 2016
2^{2}×83 Posts 
Primality test based on factorization of n^2+n+1
The primality tests for n if n^21 is factored 33.33% or higher are found here. What about the neoclassical tests for n? Suppose that n^31 is factored 33.33% or higher. Is there a way to prove n prime WITHOUT using ECPP? For example, is it possible to prove this prp (n) prime given the known factorization of n^31? I would supposed the neoclassical tests would be used, but I'm not sure. Any ideas? Thanks.

20180204, 21:35  #2 
"Rashid Naimi"
Oct 2015
Remote to Here/There
5×439 Posts 
If you know all the prime factors of n^31, then you know all the factors of n1.
This is because n1  n^31. So you can use Lucas primality test to prove that n is prime. If the partial know factors of n^31 are exclusive to n1 factors then it would seem intuitively irrelevant to the primality of n, unless you can find an association between the (extra) factors of n^m1 and n. I doubt such association exits. Last fiddled with by a1call on 20180204 at 21:48 
20180204, 21:59  #3 
"Rashid Naimi"
Oct 2015
Remote to Here/There
4223_{8} Posts 
I could not find your reference to primality testing of n when n^21 is 1/3 factored.
https://primes.utm.edu/prove/index.html Please provide a direct link. 
20180205, 00:20  #4  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×7×461 Posts 
How about Williams, 1978 reference here?
Quote:


20180205, 05:19  #5  
"Sam"
Nov 2016
2^{2}·83 Posts 
Quote:
first choose a (proven) prime p = 1 (mod 3) and let a be a primitive root mod p. compute m = a^((p1)/3) mod p If this m is (probably) prime, then most likely it can be proved using the fact that p is roughly the size of m (if not greater), and p divides m^31. If this m is not prime, then choose some other prime q (can be small), and try and see if m = a^((q1)*((p1)/3)) mod (q*p) is (probably) prime. If so, then no need to continue further. If not try and find the smallest k such m = a^((q1)*q^(k1)*((p1)/3)) mod (q^k*p) is (probably) prime. For a quick demonstation of a another probable prime n (which may be proved prime based on the factorization of n^31 (and more specifically n^2+n+1). First we have p = 13 and 6 a primitive root mod p (a = 19). We find that 19^((p1)/3) mod p isn't prime so we choose another prime q = 7. Jumping ahead to step 2 (plus a little more work applied) we find that m = a^(4*7^321) mod (7^322*13) is prp (already shown prime) and m^2+m+1 is 50% factored (33.33% for m^31). A proof based on factorization of m^2+m+1 would be easy to do for this m. Last fiddled with by carpetpool on 20180205 at 05:22 

20180205, 05:20  #6 
"Sam"
Nov 2016
332_{10} Posts 
@a1call:
https://primes.utm.edu/prove/prove3_1.html https://primes.utm.edu/prove/prove3_2.html https://primes.utm.edu/prove/prove3_3.html Hope you find these useful. 
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