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 2018-10-27, 05:37 #12 paulunderwood     Sep 2002 Database er0rr 3·1,049 Posts I have written up a refinement of this resultant idea, proof lacking -- anyone? http://www.worldofprimes.co.uk/resul...robenius/index
 2018-11-09, 07:07 #13 paulunderwood     Sep 2002 Database er0rr 314710 Posts 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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