20181206, 19:26  #1 
Mar 2016
2×3^{3}×5 Posts 
(a+bi)^p, prime test in C
A peaceful evening,
i think prime test similear to the little Fermat may be possible in the complex plane, Are there some literature of pdfs to this topic availible ? Greetings Bernhard 
20181206, 22:14  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9,127 Posts 

20181228, 19:19  #3 
Mar 2016
2×3^{3}×5 Posts 
I am too stupid to find some prime tests with complex numbers,
or prime tests with adjoined square roots from google or wikipidia. But as this mathematical subject seems to be interesting for other mathematicians, i would repush this thread. As (a+bI)^p = (abI) mod p; for p=3 mod 4 otherwise (a+bI)^p = a+bI mod p for p=1 mod 4 is right for p element P; a, b element N Or A=sqrt(2) As (a+bA)^p = (a^p+(b^p)A^p) mod p is right for p element P These tests with two variable seem to me stronger than the fermat test with one variable. I thought there must be some mathematicians who have treated this subject already. Greetings from the primetests Bernhard @Batalov, if the day is really bad and sad, try to throw a look of the perspective of a small and blue smurf, some times the change of the perspective can help to get another feeling toward the subject. 
20181228, 19:28  #4  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Quote:


20181228, 20:29  #5  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9,127 Posts 
Quote:
It is obvious  to anyone with a pencil, a sheet of paper and a few minutes to spare. Just write down (a+b*I)^p = a^p + b^p * I^p =?= a + b * I^p (mod p)... <==> a^p==a (mod p) and b^p==b (mod p) So, "These tests with two variable Would you please think for just a minute before you write? 

20181229, 17:57  #6 
Mar 2016
2·3^{3}·5 Posts 
Let us see, where the misunderstanding comes from:
1. You assume that p is a prime 2. You simplify the equation (The missings binomialcoefficients are equal 0, if p is prime 3. You conclude that two singles fermat tests are the same. For a prime test i assume 1. that p is either prime or not 2. if p is not a prime, the complex test could not be simplify From the mathematical point there is a difference between our assuming conditions. The statement that the complex test is stronger than the fermat test bases on the number of pseudoprimes which pass the test. And you are right that the complex test is more expensive than two Fermat test, by the way i think it belong to three Selfridges. Normally i think about the topics i deal and sleep a night about it. Greetings from the complex plane Bernhard Last fiddled with by bhelmes on 20181229 at 18:04 
20181229, 19:23  #7  
Sep 2002
Database er0rr
D56_{16} Posts 
Quote:
It is 2 Selfridges. For a list of candidates it is way more efficient to do an aPRP test, then a bPRP test. Last fiddled with by paulunderwood on 20181229 at 19:24 

20181229, 20:28  #8 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9,127 Posts 

20181230, 17:21  #9 
Mar 2016
2·3^{3}·5 Posts 
Nice to see you again in this conversation, i think that you could solve some problems about this topic. 1. Let p=3 mod 4: (a+bI)^(p²1)=1 mod p; if p element P (a+bI)^p = (abI) mod p Is the second test a p1 or a p+1 test ? 2. Let A=sqrt(2): (a+bA)^p = a+b(A^p) mod p; if p element P Is this test stronger than the equivalent in the complex plane ? (I think the binominalcoefficients are symmetric and therefore it migth be better to weight one variable) 3. Let c^[(p1)/2]=p1 mod p (a non quadratic residue) Then (a+b(sqrt(c))^p=a+b[sqrt(c)] mod p; if p element P I think this strengthen the test of 2. Greetings from the primitive roots Bernhard 
20190103, 17:44  #10 
Sep 2002
Database er0rr
2×3×569 Posts 
Serge gave you a clue with "=?=" above. You can just do 1 selfridge tests as the result of the binomial expansion over p.

20190103, 23:51  #11 
Mar 2016
2×3^{3}×5 Posts 
A peaceful night for you,
I think (a+bI)^p=abI mod p for p=3 mod 4 is a p+1 test: For example (2+11I)^16=1 mod 31 Greetings from the complex plane Bernhard 
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