chongo@nsc.UUCP (Landon Noll) (03/02/84)
for folks who STILL think that Fermat's Last Therom is solved, factoring and finding primes is trivial, and the RSA system is cracked wide open: the New Scientist had an article debunking Arnold Arnold. it mentioned that Nelson Stephens at Cardiff had factored some number that AA claimed was prime. no proof of this factoring was given. Arnold Arnold claims that 2^2143138339-1 is prime by his wonderful new method. Below is a proof that 295753090783 is a factor of that number. this shows that Arnold Arnold's method is bogus, and thus Arnold Arnold himself is bogus. <-- Non math folks can Q now --> Given: p = 2143138339 NOTE: p (base 2) is 1111111101111011011001000100011 k = 69 f = 2*k*p+1 = 295753090783 Prove that f is a factor of 2^p-1 Proof: We shall show that 1 = 2^p (mod f) clearly 2^0 (mod f) = 1 so 2^1 (mod f) = 2 Note: (2^p (mod f))^2 = 2^(p+p) (mod f) so 2^3 (mod f) = 8 Note: 2*(2^p (mod f))^2 = 2^(p+p+1) (mod f) so 2^7 (mod f) = 128 so 2^15 (mod f) = 32768 so 2^31 (mod f) = 2147483648 so 2^63 (mod f) = 177029235526 so 2^127 (mod f) = 192020617988 so 2^255 (mod f) = 239465908393 so 2^510 (mod f) = 211468983839 so 2^1021 (mod f) = 238984331980 so 2^2043 (mod f) = 8164330261 so 2^4087 (mod f) = 25364931418 so 2^8175 (mod f) = 21944551404 so 2^16350 (mod f) = 289452419893 so 2^32701 (mod f) = 280695646176 so 2^65403 (mod f) = 284540782241 so 2^130806 (mod f) = 126368760506 so 2^261613 (mod f) = 191444771271 so 2^523227 (mod f) = 264589510491 so 2^1046454 (mod f) = 35976998341 so 2^2092908 (mod f) = 192748064192 so 2^4185817 (mod f) = 138895788023 so 2^8371634 (mod f) = 87718855011 so 2^16743268 (mod f) = 73957365414 so 2^33486536 (mod f) = 182566897401 so 2^66973073 (mod f) = 95112577885 so 2^133946146 (mod f) = 121160121580 so 2^267892292 (mod f) = 87705983429 so 2^535784584 (mod f) = 171686937707 so 2^1071569169 (mod f) = 236034530131 so 2^2143138339 (mod f) = 1 and as a result: 2^2143138339-1 (mod f) = 0 therefore f is a factor of 2^p-1 chongo <next?> /\Mp/\