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2008-05-06, 02:50   #2
gd_barnes

May 2007
Kansas; USA

237578 Posts

Quote:
 Originally Posted by Siemelink 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^n-1 = (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.

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