20170205, 02:45  #1 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2006_{10} Posts 
LL Type Test for the Form 2n3
Hi,
Is there a LL type test for numbers of the form n^23? If so is it as efficient? Thanks in advance. 
20170205, 02:59  #2  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Quote:
n^23 = 6y+1 leads to n^2 = 6y+4 which if you test squares will get you n=2,4 mod 6 n^23 = 6y+5 leads to n^2=6(y+1)+2 which leads to no possible result as n^2 can only be 0,1,4, or 3 mod 6 

20170205, 03:08  #3 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2×17×59 Posts 

20170205, 03:32  #4 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
I think we can put it in the correct form to do one of the tests listed here ( even if it might take Proth's theorem). but I don't know if it would be efficient.

20170205, 18:40  #5 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
41·229 Posts 

20170205, 19:05  #6 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2×17×59 Posts 
Thank you.
Just the sort of answer I appreciate for Yes or No questions. Last fiddled with by a1call on 20170205 at 19:06 
20170205, 20:43  #7 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
really not even LLR ? I would get something like (2^n1)*2^1+1 for example though I guess that's too simple an attempt at an actual answer. edit: doh idiocy alert (2^n1)*(2^1)1 which makes it a proth theorem prime according the LLR wikipedia.
Last fiddled with by science_man_88 on 20170205 at 21:18 
20170205, 21:36  #8  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
22255_{8} Posts 
Quote:
What is the relationship between k and 2^{n} for the Proth theorem to be applicable? 

20170205, 21:40  #9  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20170205, 22:14  #10  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
22255_{8} Posts 
Quote:
I am not asking you read "closely enough". I am asking to read the first line. Hint: the first line of both theorems clearly says k < 2^n, doesn't it? 

20170205, 22:18  #11 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 

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