20070314, 22:00  #1 
Feb 2005
2^{2}·3^{2}·7 Posts 
primality of ((p+1)^p1)/p^2
Is there any fast method (or software) to prove primality of numbers of the form N(p) = ((p+1)^p1)/p^2 ?
In particular, I am interested in proving that N(4357) is prime. 
20070314, 22:05  #2 
Nov 2003
7460_{10} Posts 

20070314, 23:16  #3 
Feb 2005
374_{8} Posts 

20070315, 14:56  #4 
Sep 2002
Database er0rr
3,533 Posts 
Last fiddled with by paulunderwood on 20070315 at 15:05 
20070316, 09:31  #5 
Mar 2006
2^{2} Posts 
N(4357) is prime,and it has 15850 digits ! 
20070316, 10:45  #6 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
armstrong1,
how did you prove that? Alex 
20070317, 01:44  #7 
Mar 2006
2^{2} Posts 
oh,i compute it in maple 10 the software !

20070317, 04:36  #8 
Sep 2002
Database er0rr
3533_{10} Posts 
Do you have a reference to the socalled proving algorithm used?

20070319, 06:20  #9 
Mar 2006
2^{2} Posts 
oh, no, i just compute it !

20070319, 13:01  #10 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Are you certain that Maple used a primality proving algorithm, not just a probable primality testing algorithm?
Which Maple function specifically did you use to test primality of this number? Alex 
20070319, 14:23  #11 
Feb 2007
2^{4}×3^{3} Posts 
Conjecture: N(n)=0 mod M(n) for n=2^k1.
But N(66^2+1) You must have a powerful machine to use Maple's isprime() on that number... (on my laptop it takes too long :( !) However, no factor below 2^32... (pari's default primelimit...) 
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