20120402, 05:20  #1 
Nov 2011
2^{2}·3 Posts 
Proof of Primality Test for Fermat Numbers
Let be a Fermat number of the form :
Next , let's define sequence as : Then : Proof is attached . Any constructive comment is appreciated . 
20120402, 06:01  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×3^{2}×7×37 Posts 
Not too late for the April 1st! (at least in our TZ)

20120402, 07:24  #4 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3×29×83 Posts 
Nice link.
Going by the bottom comment on the only answer, the two tests are exactly the same, and the other test is from 1960, so... I guess you could see if your proof improves the original proof? (I don't have anywhere near the sort of knowledge to examine the proofs.) 
20120402, 07:40  #5  
Nov 2011
2^{2}×3 Posts 
Quote:


20120402, 08:02  #6 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3×29×83 Posts 
No, I'm pretty sure they're the same. R0=S0=8; Rn=R(n1)^22; Sn=(S(n1)^22)^22. Therefore, S1=R2, and trivially R2n=Sn for all n (n>=0). Your test ends with i=2^(n1)1, while the other test ends with k=2^n2=2i, so the sequences are identical. All your sequence does is the same as two iterationsof R, and then do half the iterations, but that's still exactly the same thing.
Last fiddled with by Dubslow on 20120402 at 08:05 
20120402, 08:59  #7 
Mar 2010
2^{6}×3 Posts 
Sorry i did not notice the assumption n>=2
Last fiddled with by literka on 20120402 at 09:05 
20120402, 09:03  #8 
Nov 2011
2^{2}·3 Posts 
Yes , you have missed condition :

20120402, 09:13  #9 
"Tapio Rajala"
Feb 2010
Finland
3^{2}·5·7 Posts 
As underlined by Dubslow, the most satisfactory answer one can give is already given in Emil Jeřábek's comment (behind the link I posted). The test is the same as Inkeri's. So, nothing new.

20120402, 11:40  #10 
Mar 2010
2^{6}·3 Posts 
I noticed this shortly after uploading post. I cannot compare your work with work of Inkeri, which I do not know. I have few editorial remarks, which may be useful for you. Everywhere you use sign of equivalence but in Lemma 2.1. you use sign of equality. I would, in your place, specify the range of x, y, in Lemma 2.1., since you use Lemma 2.1. for noninteger elements. 
20120402, 12:31  #11  
Nov 2011
2^{2}·3 Posts 
Quote:


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