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Old 2011-05-19, 21:09   #11
ATH
Einyen
 
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Dec 2003
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Quote:
Originally Posted by ATH View Post
You can simplify this alot:
k=2n-1, m=k*(n-1)-n+2, x=m*k
Now your conditions is equivalent to: 2x-1 = 1 (mod k) AND (k-n+2)x-1 = 1 (mod k)
Actually only the condition (k-n+2)x-1 = 1 (mod k) is needed, I get the same results just searching for numbers fulfilling this.

Last fiddled with by ATH on 2011-05-19 at 21:09
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