20120330, 11:37  #1 
Feb 2012
Paris, France
5×31 Posts 
Primes of the form a^(2^n)+b^(2^n)
Let's consider numbers of the form a^(2^n)+b^(2^n)
where (a,b) is a couple of integers such that a>b>1 and gcd(a,b)=1 and n integer >= 1. Is there a primality test which could be applied to such numbers? I've done a search on the internet but I did not find an answer, I was able to get my hands on a paper by Anders Björn and Hans Riesel but it only deals with factors of such number (for a,b <= 12). Example: 150^2048+7^2048 is PRP Code:
> pfgw64 f q"150^2048+7^2048" PFGW Version 3.6.2.64BIT.20120309.Win_Dev [GWNUM 27.4] 150^2048+7^2048 is 3PRP! (1.5978s+0.3911s) to produce a primality certificate for that 4457 digits number. Was it the only way I can prove the number prime? Has anybody out there been interested in this form of numbers? Last fiddled with by YuL on 20120330 at 12:01 
20120330, 12:51  #2 
Romulan Interpreter
Jun 2011
Thailand
2·13·337 Posts 
Isn't this a subset of GF(a,b)? And you wouldn't need to go so high to make the point: 3^2+2^2=13, prime, 2^4+3^4=97 prime, etc.

20120330, 12:52  #3 
"William"
May 2003
New Haven
2·3^{2}·131 Posts 

20120330, 13:11  #4  
Feb 2012
Paris, France
9B_{16} Posts 
Quote:
<< There are two different definitions of generalized Fermat numbers, one of which is more general than the other. Ribenboim (1996, pp. 89 and 359360) defines a generalized Fermat number as a number of the form a^(2^n)+1 with a>2, while Riesel (1994) further generalizes, defining it to be a number of the form a^(2^n)+b^(2^n) >> Last fiddled with by YuL on 20120330 at 13:13 

20120330, 13:31  #5 
Nov 2003
2^{2}×5×373 Posts 
AKS, APRCL, ECPP come to mind.
Last fiddled with by R.D. Silverman on 20120330 at 13:31 
20120330, 15:26  #6 
Feb 2012
Paris, France
10011011_{2} Posts 
OK, so this basically confirms that ECPP using Primo is the way to go, I've already proven a few:
8100^1024+203^1024 78^2048+41^2048 150^2048+7^2048 53^4096+2^4096 but 72^8192+43^8192 is 15216 digits long, I wonder how long would it take to produce the certificate, based on the ones I have already done I estimated it would take 4 months, I'm still thinking whether I will start the fight against that behemoth... 
20120330, 15:55  #7 
(loop (#_fork))
Feb 2006
Cambridge, England
2·3,181 Posts 
These look rather random numbers; are they the smallest values of a+b with a^2048+b^2048 prime or something?

20120330, 18:29  #8  
Einyen
Dec 2003
Denmark
5600_{8} Posts 
Quote:
http://www.mersenneforum.org/showthread.php?t=10761 Cybertronic seems like an expert on Primo, but he is not just starting it and letting it run, he is doing it a lot of manual work running one primo on each core and then combining the proof somehow, which he tries to explain in that thread. He did 11,467 digits in 5months and 8,724 digits in 1030hours ~ 43 days, and he says runtime follows digits^3.75, so 15,216 should be like 1415months, and he also estimates 17,000 digits at 18months. 

20120330, 18:37  #9 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9,127 Posts 
Primo on linux is already doing all the Luhnification on N threads. No need for manual interventions anymore.
Also, consider all already known GFNs (e.g. http://www.primenumbers.net/prptop/s...096%2Bb%5E4096 off the top of my head ...and there are other sites, of course) before searching for more of these PRPs. 
20120330, 19:07  #10 
Mar 2010
2^{6}·3 Posts 
Of course, every prime factor of a Fermat number is of the form a^2+b^2. So, for n=1.
I know only one factor of a Fermat number of the form a^4+b^4 (i.e. for n=2). It is 641=5^4+2^4  factor of F5. I know only one prime factor of Fermat number of the form a^4+b^2. Factor of F18: 13631489=5^4+3692^2 Last fiddled with by literka on 20120330 at 19:07 
20120330, 20:12  #11  
Feb 2012
Paris, France
9B_{16} Posts 
Quote:
 for n=11 the smallest prime is 22^2048+3^2048<2750 digits> the next one being 43^2048+2^2048<3346 digits>  for n=11 the smallest prime such that either a or b =7 is 150^2048+7^2048<4457 digits>  for n=12 the smallest prime is 53^4096+2^4096<7063 digits> and (surprisingly) the next one also has a=53: 53^4096+48^4096<7063 digits> (actually the latter is PRP and the ECPP proof is in progress). 

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