2019-10-09, 19:46 | #1 |
Jun 2019
3^{2} Posts |
Condition on composite numbers easily factored (Challenger)
q prime numbre
p prime numbre and q>p lets c = qp e= 2^p mod q if we know e we can factore (c) ok letz try with c and e Last fiddled with by baih on 2019-10-09 at 19:51 |
2019-10-10, 11:14 | #2 |
"Robert Gerbicz"
Oct 2005
Hungary
2×3×223 Posts |
With "a mod n" we denote the smallest non-negative residue modulo n.
We're given e=2^p mod q and calculate (in polynomial time) r=2^n mod n What is r mod q ? Easy: r mod q=(2^n mod n) mod q=2^n mod q=(2^q)^p mod q=2^p mod q=e (used Fermat's little theorem). So we know that r mod q=e, hence r-e is divisible by q, so q | gcd(r-e,n), and if this is not n, then we could factorize: because q=gcd(r-e,n) and with a division p=n/q. The (only) hole in this proof is that (in rare cases) it is possible that n=gcd(r-e,n), for example this is the case for n=341. To see that it is really working: Code:
f(n,e)={r=lift(Mod(2,n)^n);d=gcd(r-e,n);return(d)} p=nextprime(random(2^30)); q=nextprime(random(2^30)); n=p*q;e=lift(Mod(2,q)^p); print([p,q]); f(n,e) (one possible output) ? [623981947, 805922797] ? %42 = 805922797 Last fiddled with by R. Gerbicz on 2019-10-10 at 11:16 Reason: typo |
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