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Old 2018-10-27, 05:37   #12
paulunderwood
 
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I have written up a refinement of this resultant idea, proof lacking -- anyone?

http://www.worldofprimes.co.uk/resul...robenius/index
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Old 2018-11-09, 07:07   #13
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Default Semi-primes again

Recapping the definitions:

n(b)=\prod_{i\le s}(k_i b+1)^{e_i}

gcd(k_1,\ldots,k_s)=1

f(b)=n(b)+1

g(b)=\prod_{i\le s}(k_i b+2)^{e_i}

H=gcd(f(b),polresultant(f,g))

Conjecture: x^{n+1}\equiv 1 \pmod{n, x^2-ax+1} implies

x^H\equiv 1 \pmod{n, x^2-ax+1} (**) where

Jacobi(a^2-4,n)==-1.

Now let the semi-prime n=(Sb+1)(Tb+1)

I claim that if (**) holds then:

\gcd((\sum_{i=0}^{\lfloor\frac{S-1}{2}\rfloor}-1^i{S-1-i\choose i}a^{S-1-2i})(\sum_{j=0}^{\lfloor\frac{T-1}{2}\rfloor}-1^j{T-1-j\choose j}a^{T-1-2j}),n) = n

Last fiddled with by paulunderwood on 2018-11-09 at 07:16
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