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

CC BY-SA 4.0
2019-02-15T03:48:26.527Z
Is it possible to learn this power?
Pieter Wuille
CC BY-SA 4.0
2019-02-15T06:10:35.090Z
I'm afraid you might not be random enough.
G. Maxwell
CC BY-SA 4.0
2019-02-15T03:46:46.087Z
2019-02-15T03:46:46.087Z
G. Maxwell
<p>if in a curve y2 mod p = x3+ 7 mod p ( ecdsa-secp256k1)</p>

<p>p=115792089237316195423570985008687907853269984665640564039457584007908834671663</p>

<p>G=55066263022277343669578718895168534326250603453777594175500187360389116729240,32670510020758816978083085130507043184471273380659243275938904335757337482424</p>

<p>nG=21505829891763648114329055987619236494102133314575206970830385799158076338148,98003708678762621233683240503080860129026887322874138805529884920309963580118</p>

<p>2nG=72488970228380509287422715226575535698893157273063074627791787432852706183111,62070622898698443831883535403436258712770888294397026493185421712108624767191</p>

<p>How do i find 'n' ?</p>

CC BY-SA 4.0
2019-02-14T09:13:38.823Z
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
CC BY-SA 4.0
2019-02-15T03:57:28.580Z
Correction: solving this problem *when n is uniformly randomly chosen* is assumed to be infeasible.
Pieter Wuille
<wallet><ethereum><ecdsa><secp256k1>
Generator Point G multiple
James
# Generator Point G multiple

(Asked by: username_0 on 2019-02-14)

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' ?

**Tags:** wallet, ethereum, ecdsa, secp256k1

> **Comments:**
> **username_1**: This requires solving the discrete logarithm problem, which is assumed to be infeasible. If it wasn't, elliptic curve cryptography would be insecure.  
> **username_1**: Correction: solving this problem *when n is uniformly randomly chosen* is assumed to be infeasible.  

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## 1 answers

### Answer by username_2 on 2019-02-15. Score: 3

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

> **Comments:**
> **username_1**: Is it possible to learn this power?  
> **username_2**: I'm afraid you might not be random enough.  

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