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Old 2011-05-19, 19:19   #6
ATH
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Dec 2003
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Testing natural numbers from 2 to 11213, this conjecture is true only for the exponents of the 23 first prime mersenne numbers except 2 is missing and 4 is included:
3,4,5,7,13,17,19,31,61,89,107,127,521,607,1279,2203,2281,3217,4253,4423,9689,9941,11213

Quote:
Numbers n such that a==0(mod k) and b==0(mod k), where k=2^n-1, m=(2^n-1)*(n-1)-n+2, x=m*(2^n-1), 2^(x-1)==(a+1)(mod x), m^(x-1)==(b+1)(mod x)
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)

Explanation:
Since x=m*k:
2^(x-1)==(a+1)(mod m*k) => 2^(x-1) = m*k*c1 + a+1 for some constant c1
and
a==0(mod k) => a = k*c2 for some constant c2.

Inserting k*c2 for a in the first equation gives:
2^(x-1) = m*k*c1 + k*c2 + 1 = 1 (mod k)

In the same way it can be shown:
m^(x-1) = 1 (mod k)
But instead of m it's faster to use m mod k:
m = k*(n-1)-n+2 = -n+2 (mod k) = k-n+2 (mod k) (-n+2 is negative, so if we want we can add k to make it positive)
So:
m^(x-1) = (-n+2)^(x-1) = (k-n+2)^(x-1) = 1 (mod k)

This is why n=2 doesn't work since k-n+2 = k and k^(x-1) = 0 (mod k)

Last fiddled with by ATH on 2011-05-19 at 19:26
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