20200318, 14:35  #45  
Nov 2003
2^{2}·5·373 Posts 
Quote:
When you say "consequently", the obvious reply is "consequently to what?". State mathematically what condition you mean. Math is not "casual English". Do you not know what "order of an element [in a group]" means? It is a fundamental concept that you need to learn. And you also need to learn Lagrange's Theorem if you want to continue further. 

20200318, 14:47  #46  
Nov 2003
2^{2}×5×373 Posts 
Quote:
incorrect mathematical statements. 

20200318, 15:04  #47 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{4}·3^{3}·13 Posts 
In the spirit of this thread I also want to write an HLL version of the the LL test. Why not? Following is some awesome python code. And as a bonus it also does a PRP test.
Code:
for p in range(3, 1280, 2): mp, s = pow(2, p)  1, 4 for i in range(2, p): s = (s * s  2) % mp if s == 0: print("2^{}1 is prime".format(p)) if pow(3, mp + 1, mp) == 9: print("2^{}1 is PRP".format(p)) 
20200318, 15:41  #48  
Sep 2002
Database er0rr
3335_{10} Posts 
Quote:


20200318, 15:58  #49  
Nov 2017
Karlsruhe, Germany
31 Posts 
Quote:
\(S \left( 1 \right) = 4 \\ S \left( n + 1 \right) = {S \left( n \right)}^2  2 \\ M_p = 2^p  1 \\ n < p, n \in \mathbb{N}, p \in \mathbb{P}\) 

20200318, 17:19  #50  
Sep 2002
Database er0rr
5×23×29 Posts 
Quote:
If S_k^22 == 1 mod M_p would imply S_k^2 == 3 mod M_p, but 3 is a QNR over M_p for all prime p>2. Proof? Last fiddled with by paulunderwood on 20200318 at 17:23 

20200318, 17:23  #51 
Bamboozled!
May 2003
Down not across
2^{3}·31·41 Posts 
One error you have just made is to ignore the second and third options in my list. To jog your memory they werre "naivety" and "whatever".
Naivety is both inevitable and forgivable^{*}; everyone is naive in any specific field until they learn not to be. Your naivety (in my opinion, others may differ) is in assuming that mathematicians invariably tolerate sloppy language when used to describe a problem which can be described precisely. Inherent in that sloppiness is a failure to distinguish clearly between the result of operators in programming languages and the mathematical concept of an equivalence class. "Whatever" is harder to characterize. One thing in your favour is your adherence to one of the laws of optimization: First get it right, then get it fast. Perhaps if you had stated this upfront your post may have had a different outcome. Very likely so. Not everyone is polite.  * I've been getting all sorts of since I joined Twitter because my Asperger's makes it difficult for me to read subtexts in tweets from neurotypicals. I have been trying hard to learn and I believe that I'm improving but it still trips me up all too often. 
20200318, 17:59  #52 
Nov 2003
2^{2}·5·373 Posts 
Not if M_p is prime. If M_p is composite then 4 can generate a subgroup.

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