20101227, 17:52  #1 
Dec 2010
74_{10} Posts 
Primorials squared primes?
Is there any sort of search for primes of the form (2*3*5*7*...*p)^2 + 1?
I noticed that projects search for both factorial and primorial primes, but I'm curious about squaring the primorial first, then adding one. Numbers of the form n^2 + 1 are only divisible by primes congruent to 1 mod 4. Wouldn't this imply that they are more frequently prime than the average number? Additionally, wouldn't it be easier to trial factor such numbers? How about a modified "primorial" involving only two and the primes congruent to 1 mod 4? In other words, (2*5)^2 + 1 = 101 (a prime), (2*5*13)^2 + 1 = 16901 (a prime), (2*5*13*17)^2 + 1 = 4884101 (a prime), and so on. 
20101227, 18:27  #2  
Aug 2006
1011100100101_{2} Posts 
Quote:
chance for p#^2 + 1 to be prime, so there should be something like primes up to x#^2 + 1. Quote:
You could trial factor almost twice as quickly, so factoring to (say) 40 bits would only take a little longer than factoring a 'normal' number to 39 bits. Last fiddled with by CRGreathouse on 20101227 at 18:59 

20101227, 18:49  #3 
Jun 2003
11131_{8} Posts 

20101227, 18:59  #4 
Aug 2006
3×5^{2}×79 Posts 

20101227, 19:42  #5  
"Gang aft agley"
Sep 2002
EAA_{16} Posts 
Quote:


20101228, 07:23  #6 
"Gang aft agley"
Sep 2002
2×1,877 Posts 
I understand that you intend n to be even here. Although these generated numbers may be divisible by primes congruent to 1 mod 4, might they also be divisible by a even number of primes congruent to 1 mod 4?

20101228, 15:17  #7 
Jun 2003
7·11·61 Posts 
No. They are GFNs  b^(2^n)+1  and hence divisibility rules for Fermat numbers apply, i.e. factors must be of the form k.2^(n+1)+1

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