20200611, 23:46  #1 
"James Short"
Mar 2019
Canada
1011_{2} Posts 
Proths of the form 2^{k+1}(2^{k}  1) + 1, 2k+1 is prime
(I guess am sort of answering my own question here......but here goes)
Are Proth Primes of the form worthwhile to search for? Fyi  I'm specifically referring to the special case when is itself a prime number and when So here are my reasons for choosing these numbers and restrictions. 1) I noticed that potential composite factors have to satisfy some very strict congruence conditions when is prime (even more strict than prime factors of Mersenne numbers). For example, because the order of 2 over can be shown to be equal to . 2) Sieving out composite Proths of this form is very easy due to the fact that all potential prime factors have the form (unless , but the congruence restrictions on prevent this). Fyi  I personally used the Pollardrho algorithm with the iterating function to weed out many composites (here is a basic program: rho1(n)= { local(x,y); x=2; y=2^(24p) + 1; while(gcd(yx,n)==1, x=(x^(24p)+1)%n; y=(y^(24p)+1)%n; y=(y^(24p)+1)%n ); gcd(n,yx) } ) Note: You don't necessarily need to use the 'extra' multiplier in the exponent of the iterating function, however the exponent of the iterating function definitely needs to be a "highly composite number" multiple of (if not equal to itself) to be the most efficient at weeding out composite Proths of this form . Alternatively, the p1 test could be used instead to weed out composites. I'm not sure which of the two would be more efficient tbh (if anyone knows the answer to this, that'd be awesome!). 
20200612, 05:34  #2 
"Sam"
Nov 2016
311 Posts 
I think you are referring to Gaussian Mersenne norms.

20200615, 17:42  #3  
"Jeppe"
Jan 2016
Denmark
10011011_{2} Posts 
Quote:


20200615, 19:07  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×3^{3}×13^{2} Posts 

20200615, 21:00  #5  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·3^{3}·13^{2} Posts 
Quote:
* n needs to be prime  why? because otherwise there is an algebraic factorization. (And in the algebraic factorization you will find both + and  forms) * Any factor (except 5 which is intrinsic) is of form 2 * k * p + 1, so trial factoring is simplified but is very similar to Mersenne project: you don't sieve databasewide  you prefactor each candidate. Furthermore, just like factors of Mersenne composites, these will not share the same factors (except the trivial case when one divides the other). ...that's just a few quick thoughts is a spare time during lunch break. P.S. All primes of this form at this time are known  loosely speaking to the limit of p < 10,000,000. So searching below this limit will not bring too much joy to the searcher. 

20200620, 07:55  #6 
"Jeppe"
Jan 2016
Denmark
5×31 Posts 
If you consider the numbers 2^(4k+2) + 1, there are two kinds of algebraic factorizations for them. One is because 4k+2 = 2(2k+1), so the exponent has an odd divisor, therefore:
2^(4k+2) + 1 = (2^2)^(2k+1) + 1 = (2^2 + 1)[(2^2)^(2k)  (2^2)^(2k1) + ...  (2^2)^3 + (2^2)^2  2^2 + 1] But this simply says that 2^2 + 1 = 5 divides 2^(4k+2) + 1 which is also quite obvious without the factorization above. Set n=2k+1. When n is a prime, there are no other nontrivial odd factors of the exponent 4k+2 than n, and then we do not get more factorizations of the above type. However, Aurifeuille noted another algebraic factorization of 2^(4k+2) + 1, namely 2^(4k+2) + 1 = (2^(2k+1)  2^(k+1) + 1)(2^(2k+1) + 2^(k+1) + 1) For each k, one of the two factors here is necessarily divisible by 5. Then the other may be prime. We can also write Aurifeuille's factorization with j = k+1 instead: 2^(4j2) + 1 = (2^(2j1)  2^j + 1)(2^(2j1) + 2^j + 1) or with n (which is odd): 2^(2n) + 1 = (2^n  2^((n+1)/2) + 1)(2^n + 2^((n+1)/2) + 1) Let n be this odd number. The relation to Mersenne primes, as explained on Caldwell's page linked by carpetpool and Batalov above, is that the complex number (1+i)^n  1 where i is a square root of minus one, is a prime in the ring of Gaussian integers exactly if the one of Aurifeuille's factors which is not a multiple of 5, is an ordinary prime. The classical (https://en.wikipedia.org/wiki/Aurife...zation#History) example is n = 29 (so k=14 and j=15) where Aurifeuille says: 2^58 + 1 = (2^29  2^15 + 1)(2^29 + 2^15 + 1) where the latter factor is prime (so a Gaussian Mersenne norm prime corresponding to the Gaussian prime (1+i)^29  1 where i is either of the two square roots of minus one). /JeppeSN 
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