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Old 2008-04-13, 19:45   #1
Siemelink
 
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Jan 2006
Hungary

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Default 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.
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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.
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