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Old 2021-03-06, 18:19   #954
warachwe
 
Aug 2020

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Quote:
Originally Posted by henryzz View Post
I think the trick would be to prove that:
2^(2^3 * 3^2 * 5 * 7 * k + 1) - 1 == 1 mod d
3^(2^3 * 3^2 * 5 * 7 * k + 1) - 1) / 2 == 1 mod d
6^(2^3 * 3^2 * 5 * 7 * k) == 1 mod d
For p prime, and and a such that gcd(a,p)=1, we must have a^(p-1) == 1 mod p.
So if p-1 divides 2^3 * 3^2 * 5 * 7, it means 2^(2^3 * 3^2 * 5 * 7 * k) == 1^k == 1 mod p, hence 2^(2^3 * 3^2 * 5 * 7 * k+1)-1 == 1^k*2-1 == 1 mod p.
Same thing for 3 and 6.

That take care of p=5+,7+,11,13,19,29,31,37,41,43,61,71,73,127,181,211,281,421,631
(+ This only prove that 5,7 divide s(6^(2^3 * 3^2 * 5 * 7 * k)), not 5^2,7^2)
For p=337, we have (p-1)/2 divides 2^3 * 3^2 * 5 * 7. Possible value for a^((p-1)/2) is +-1 mod p. I guess that mean 2^((337-1)/2),3^((337-1)/2),6^((337-1)/2) are happened to be 1 mod 337.

To prove that 5^2,7^2 divide s(6^(2^3 * 3^2 * 5 * 7 * k)), we use the fact that if a == 1 mod p, then a^p == 1 mod p^2.
(5-1) divides 2^3 * 3^2 * 7, so a^(2^3 * 3^2 * 5 * 7) == 1 mod 5^2 for (a,p)=1
(7-1) divides 2^3 * 3^2 * 5, so a^(2^3 * 3^2 * 5 * 7) == 1 mod 7^2 for (a,p)=1
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