Jane asked for a demonstration when ric.judor wrote:> > i can find a factor of any composit numbers that satisfy n =p*q

One 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:

> 2203353588507632689540389884099234780425798106022737554148425561277092513841358286596173426627853446495163308636257511927689252306411

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

sqrt(8)N^(1/4).

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?

Peter

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[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278