20210729, 18:39  #12 
Jun 2021
3×17 Posts 
How does the refrigerator work?
Grmmmmmmmmmm Old joke. b=mod(u^n,p); a=mod(b^n,p) b^2a is a multiple of p for any integer n>0, I don't know the behavior of epsilon for every n, but seems that algorithm work only if n=2. 
20210729, 18:48  #13 
Apr 2020
3^{2}·5·19 Posts 
Try going through the same calculations that I did, but with n=3 and the square roots replaced with cube roots. Can you see where things change?

20210729, 19:24  #14 
Jun 2021
3×17 Posts 
) we start from (by)^2? For cube, (by)^3 must be expanded (b^33*b^2*y+3*b*y^2y^3) and solved in whole by the same manner, without any simplification from the start. And I'm stuck with imaginary parts and 3roots)) At the same time, cube and highger have solution, just like square!

20210729, 19:55  #15  
Apr 2020
1101010111_{2} Posts 
Going back to this:
Quote:
What is the question for higher n? Is there even a question? 

20210730, 10:30  #16 
Jun 2021
110011_{2} Posts 
Once again, it's heuristic! There need a genius to shed the light why it work!
For higher orders, I have no working code. This is not mean that is impossible. 
20210730, 13:14  #17  
Apr 2020
855_{10} Posts 
Quote:
You've got the code, you can check whether the heuristic fits the numerical evidence. If you solve my exercise about the expected descent rate from a few posts back, that will give you a hypothesis to test. 

20210730, 18:22  #18 
Jun 2021
3×17 Posts 
First, I dont know whether its well known or new. Second  the spirit of this place! And third, connected with second, plenty of unborn ideas glimpse close to our sight, here I can catch them. Thank You very much for your epsilon!

20210730, 23:11  #19  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{4}·3^{3}·23 Posts 
Refrigerator can easily do 'Grmmmmmmmmmm' and not work actually.
Quote:
Quote:


20210731, 02:13  #20  
Feb 2017
Nowhere
2·29·103 Posts 
Quote:
Another issue arises if n > 2, particularly for odd n > 2. Even if p is prime, if gcd(n, p1) = 1, every residue mod p is an n^{th} power residue. For n = 3, this is the case for every prime p congruent to 2 (mod 3). 

20210731, 15:19  #21  
Jun 2021
3×17 Posts 
Quote:
No. Talk was about quadratric case; yes for higher orders. Quote:
if b<sqrt(p) and b^3>p, a=mod(b^3,p) one roots (of 3) ((b)^3+a)^(1/3)+b (by)^4 root (b^4a)^(1/4)+b and so on. Other roots also have importance Last fiddled with by RomanM on 20210731 at 15:23 

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