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Re: The first new theorem Primes
Posted:
Sep 18, 2013 12:24 PM


can you state your premise succinctly, dood
> Since A2 = 1000000A1 > > > > A2 A1 = 999999A1 > > > > Then > > > > Q2(B)  Q1(B) = 999999A1 > > > > Dividing through by B, we obtain. > > > > Eqtn. (3a) Q2  Q1 = ( 999999A1)/B > > > > 11235053  11 = (999999)(13718)/1221 > > > > 11235042 = 11235042 > > > > We have an integer value on the left hand side of Eqtn. (3a), and thus we > > should also have an integer value on the right hand side of Eqtn. (3a). In > > order for this to be true, since A and B have no prime factor in common, > > then 999999 must be evenly divisible by B . > > > > Checking this out in the numerical example, we find. > > > > 999999/1221 = 819 > > > > An integer quotient value is produced. > > > > Thus we have shown that our randomly chosen odd integer, 1221, is a factor > > of 999999.



