Post
Topic
Board Development & Technical Discussion
Re: Pollard's kangaroo ECDLP solver
by
brainless
on 14/06/2020, 16:32:35 UTC
@mrxtraf
In the private key group (mod n) we can add, negate, and invert - this allows for multiplication and division.

In the public key group (elliptic curve mod p of size n) we can add, negate, and double only. This leads to multiplication by a scalar.

One public key corresponds to exactly one private key, and vice versa. The proof is very easy. Let G is the generator of secp256k1. Let P=k*G is a point on the curve. Let also P=k'*G. Then (k-k')*G=O => (k-k') divides n. But n is prime, hence k=k' (mod n).
That is, you can’t divide the public key by 10?
Give me any public key from which you know the private key, I will divide it by 10. And I will give in return the result in the form of a public key. And you yourself divide the private key by 10, get the public key from it and compare.

The multiplication (and division, which is multiplication with the inverse) is by scalar only. You cannot multiply two public keys without solving ECDLP first. And if you somehow can, then all coins are belong to you.

mrxtraf is saying they can "divide" the public key by 10 by multiplying the public key by the multiplicative inverse of 10 mod n.
BitCrack
apply your algo and show me pubkey div by 789 for
03D041CF467F485A96AB21EC0E1E1E26A344B28A12244320C4BDE48C123653D88F