Re: [PrimeNumbers] some composite can be broken down into key factors
Jane asked for a demonstration when ric.judor wrote:
> > i can find a factor of any composit numbers that satisfy n =p*qOne French lawyer taught us that if a number N has two factors
> > ,p>q , p-q < sqrt (p)
> Oh good. Perhaps you'd like to factor this one for us:
P and Q, which are not too far from each other, we only need
to calculate A = ceil(sqrt(N)) and get the factors immediately
as P = A + sqrt(A^2 - N) and Q = A - sqrt(A^2 - N). The "not
too far from each other" condition basically asks for the
difference of (P-Q) to be bounded (if I recall correctly) by
Now, I claim that Ric's condition of (p-q) < sqrt(p) is
actually stronger than this. If it wasn't so, we'd have
sqrt(p) >= sqrt(8)N^(1/4), which implies p >= 64*q. Then,
however p - q >= (63/64)p > sqrt(p), a contradiction.
Since your number doesn't seem to admit the factorization
described above, I believe it doesn't satisfy his inequality.
Or maybe I overlooked something?
[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278