0

if in a curve y2 mod p = x3+ 7 mod p ( ecdsa-secp256k1)

p=115792089237316195423570985008687907853269984665640564039457584007908834671663

G=55066263022277343669578718895168534326250603453777594175500187360389116729240,32670510020758816978083085130507043184471273380659243275938904335757337482424

nG=21505829891763648114329055987619236494102133314575206970830385799158076338148,98003708678762621233683240503080860129026887322874138805529884920309963580118

2nG=72488970228380509287422715226575535698893157273063074627791787432852706183111,62070622898698443831883535403436258712770888294397026493185421712108624767191

How do i find 'n' ?

  • 5
    This requires solving the discrete logarithm problem, which is assumed to be infeasible. If it wasn't, elliptic curve cryptography would be insecure. – Pieter Wuille Feb 14 at 9:13
  • Correction: solving this problem when n is uniformly randomly chosen is assumed to be infeasible. – Pieter Wuille Feb 15 at 3:57
3

You can find n by consulting a discrete log oracle, such as myself. N is 5. You're welcome.

  • 1
    Is it possible to learn this power? – Pieter Wuille Feb 15 at 3:48
  • 1
    I'm afraid you might not be random enough. – G. Maxwell Feb 15 at 6:10

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