20080413, 19:45  #1 
Jan 2006
Hungary
2^{2}×67 Posts 
base 19: some k that are square can be eliminated
Hi everyone,
I was playing today with k's that are square for base 19. I could show that k mod 10 = 4 can not have a prime, if k is a square. Could someone kindly check if what I say is logical? It's been awhile since I wrote down a proof... Thanks, Willem.  if k = m*m and n = 2p k*19^n1 = (m*19^p +1)(m*19^p 1) so for even n there is always a factor. if k mod 10 = 4 and n = 2p+1 k*19 mod 10 = 6 with 19^2p mod 10 = 1 gives k*19^(2p+1) mod 10 = 6 k*19^(2p+1)1 mod 10 = 5 so for odd n there is always a factor 5. 
20080506, 02:50  #2  
May 2007
Kansas; USA
2·5,087 Posts 
Quote:
Thanks for the astute observation Willem. This is correct. I will reflect it on the web pages and remove the appropriate k's. I'll send you a list of k's that were removed and the adjusted number of them that are remaining per the status that you posted today. Gary 

20080506, 04:01  #3 
May 2007
Kansas; USA
2·5,087 Posts 
Willem,
I found 60 k's that I inadvertantly omitted from the last update of the Riesel base 19 reservations web page. They are all in the ranges of k=219896246654 and 803364822474. I mention this because hopefully you only used my previous post to eliminate k's that were multiples of the base. If you used the web page for k's remaining, then that's not so good. They will be added back along with the removal of k's for algebraic factors plus the k's that you found primes for via your recent status report. Sorry about that...stupid parsing, cutting, and pasting of long lists of k's has bitten me more than once now. Gary Last fiddled with by gd_barnes on 20080509 at 06:49 
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