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I found a topic on bitcointalk

Public key x and y == Double(Half of the Public key x and y)

half any public key is possible how that possible , in crypto there is subtraction and multiplication only then how divition

one post Sure, just multiply the point by 2^-1 (mod n).

can some one explain more which point to multiply and what is 2^-1 (mod n)

Half of any bitcoin (crypto) public key - (public key half)

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ECC is cryptography over an elliptic curve group.

Firstly you have an elliptic curve, e.g. Bitcoin uses a Koblitz curve secp256k1 $y^2 = x^3 + 7$.

The group is defined over curve points over a finite field $F_p$ (integer modular $p$). The group elements are points on the curve. A point in the affine form consists of two coordinates $P =(x,y)$ where $x,y\in F_p$.

For group elements, you can do point addition $P+Q$, as well as scalar multiplication $sP$, where $s$ is an integer in $Z_n$ where $n$ is the order of the group (how many elements in the group).

A public key in bitcoin is a point $P$. To do $\frac{P}{2}$, you multiply $\frac{1}{2}$ to $P$ where $\frac{1}{2} $ is the multiplicative inverse of $2$ in $Z_n$. It is an integer that can be found using the extended Euclidean algorithm and is 57896044618658097711785492504343953926418782139537452191302581570759080747169 in the case of secp256k1.

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  • $\begingroup$ (at first i thank to you)private key 3 (x = f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9 y = 388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672) half of the 3 public key is (x = c62c910e502cb615a27c58512b6cc2c94f5742f76cb3d12ec993400a3695d413 y = 17f3dadd767275ddd3b23f46723631778bf01dadaebb9a953cf068712457c010) but the real output is (x = 79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 y = 483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8) 3/2 = 1 what wrong in this how to correct this $\endgroup$ – Prabu r Jun 12 '18 at 14:03
  • $\begingroup$ You are wrong in that 3/2 is not equal to 1 modulo $n$ in this context. It is $3 \cdot 2^{-1} \bmod n$. Be aware we are not using integer arithmetics, but modular arithmetics. $\endgroup$ – Changyu Dong Jun 12 '18 at 18:11

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