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^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.

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