20110401, 20:17  #1 
Dec 2009
3^{3} Posts 
The fastest primality test for Fermat numbers.
The test states that for n > 2,
F(n) is prime iff 5^((F(n)1)/4) == sqrt(F(n)1) (mod F(n)). <Thread posted on April 01, 2011. Last fiddled with by Arkadiusz on 20110401 at 20:20 
20110401, 21:35  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{4}·3·5^{2}·7 Posts 
Quote:
so you say: 5^(2^(2^n2)) mod (2^(2^n)+1) = 2^((2^n)/2) according to my research. I'll have to think on this more, I'm not that advanced. Last fiddled with by science_man_88 on 20110401 at 21:38 

20110401, 23:39  #3 
"Forget I exist"
Jul 2009
Dumbassville
2^{4}×3×5^{2}×7 Posts 

20110401, 23:50  #4 
"Forget I exist"
Jul 2009
Dumbassville
2^{4}×3×5^{2}×7 Posts 

20110402, 00:11  #5 
"Forget I exist"
Jul 2009
Dumbassville
10000011010000_{2} Posts 

20110405, 16:58  #6 
"Forget I exist"
Jul 2009
Dumbassville
2^{4}×3×5^{2}×7 Posts 
Code:
(13:56)>for(n=1,100,print1(isprime(F(n))",")) 1,1,1,1,0,0,0,0,0,0,0,0,0, *** isprime: user interrupt after 15,468 ms. (13:57)>for(n=1,100,print1(5^((F(n)1)/4)%(F(n)) == sqrt(F(n)1)",")) 0,0,1,1, *** _^_: user interrupt after 12,782 ms. it fails twice in the first 4. 
20110405, 19:39  #7  
"Forget I exist"
Jul 2009
Dumbassville
2^{4}×3×5^{2}×7 Posts 
Quote:


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