Bitcoin Forum
May 08, 2024, 04:17:01 PM *
News: Latest Bitcoin Core release: 27.0 [Torrent]
 
   Home   Help Search Login Register More  
Pages: [1]
  Print  
Author Topic: The mirror effect of public keys  (Read 208 times)
Kostelooscoin (OP)
Member
**
Offline Offline

Activity: 202
Merit: 16


View Profile
October 21, 2021, 10:11:38 AM
 #1

Does anyone know the public key mirror ?
1715185021
Hero Member
*
Offline Offline

Posts: 1715185021

View Profile Personal Message (Offline)

Ignore
1715185021
Reply with quote  #2

1715185021
Report to moderator
1715185021
Hero Member
*
Offline Offline

Posts: 1715185021

View Profile Personal Message (Offline)

Ignore
1715185021
Reply with quote  #2

1715185021
Report to moderator
1715185021
Hero Member
*
Offline Offline

Posts: 1715185021

View Profile Personal Message (Offline)

Ignore
1715185021
Reply with quote  #2

1715185021
Report to moderator
The grue lurks in the darkest places of the earth. Its favorite diet is adventurers, but its insatiable appetite is tempered by its fear of light. No grue has ever been seen by the light of day, and few have survived its fearsome jaws to tell the tale.
Advertised sites are not endorsed by the Bitcoin Forum. They may be unsafe, untrustworthy, or illegal in your jurisdiction.
1715185021
Hero Member
*
Offline Offline

Posts: 1715185021

View Profile Personal Message (Offline)

Ignore
1715185021
Reply with quote  #2

1715185021
Report to moderator
1715185021
Hero Member
*
Offline Offline

Posts: 1715185021

View Profile Personal Message (Offline)

Ignore
1715185021
Reply with quote  #2

1715185021
Report to moderator
1715185021
Hero Member
*
Offline Offline

Posts: 1715185021

View Profile Personal Message (Offline)

Ignore
1715185021
Reply with quote  #2

1715185021
Report to moderator
Kostelooscoin (OP)
Member
**
Offline Offline

Activity: 202
Merit: 16


View Profile
October 21, 2021, 10:17:54 AM
Last edit: October 21, 2021, 05:45:48 PM by achow101
 #2

here is the public key 00000000000000000000000000000000000000000000000000000000000000000001

0479be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798

and here is the mirror are fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364140

0479be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798b7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777
0379be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798

how to explain this ?



something else:

according to this shema:

0=f 1=e 2=d 3=c 4=b 5=a 9=6 8=7

look at the text in red


0479be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08   ffb10d4b8
0479be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798b7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6   f04ef2777


everything fits but the last


and

0xb7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777+0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8 = fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
a.a
Member
**
Offline Offline

Activity: 126
Merit: 36


View Profile
October 21, 2021, 11:42:14 AM
Merited by pooya87 (2), hugeblack (2), ABCbits (1)
 #3

This is the symmetry in secp256k1.

0479be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798483ada7726a3c 4655da4fbfc0e1108a8fd17b448a68554199c47d08 is the uncompressed privatekey of 1. The other one is the uncompressed privatekey of -1.

-1 => N - 1

Kostelooscoin (OP)
Member
**
Offline Offline

Activity: 202
Merit: 16


View Profile
October 21, 2021, 11:46:29 AM
 #4

ok  Wink
pooya87
Legendary
*
Offline Offline

Activity: 3444
Merit: 10555



View Profile
October 21, 2021, 02:22:43 PM
 #5

If point P is computed like P = k.G then we have -P = -k.G.
For each elliptic curve point P(x,y) we define point negation as -P(x,y) = P(x,-y) and in modular arithmetic we know that -x ≡ N-x (mod N) if 0<= x < N so both -y and -k are in congruence with N-y and N-k respectively.

.
.BLACKJACK ♠ FUN.
█████████
██████████████
████████████
█████████████████
████████████████▄▄
░█████████████▀░▀▀
██████████████████
░██████████████
████████████████
░██████████████
████████████
███████████████░██
██████████
CRYPTO CASINO &
SPORTS BETTING
▄▄███████▄▄
▄███████████████▄
███████████████████
█████████████████████
███████████████████████
█████████████████████████
█████████████████████████
█████████████████████████
███████████████████████
█████████████████████
███████████████████
▀███████████████▀
█████████
.
Kostelooscoin (OP)
Member
**
Offline Offline

Activity: 202
Merit: 16


View Profile
October 21, 2021, 04:31:27 PM
 #6

If point P is computed like P = k.G then we have -P = -k.G.
For each elliptic curve point P(x,y) we define point negation as -P(x,y) = P(x,-y) and in modular arithmetic we know that -x ≡ N-x (mod N) if 0<= x < N so both -y and -k are in congruence with N-y and N-k respectively.


ok good
NotATether
Legendary
*
Online Online

Activity: 1596
Merit: 6730


bitcoincleanup.com / bitmixlist.org


View Profile WWW
October 22, 2021, 06:12:12 AM
 #7

Also see this:

There are actually 6 combinations so only 2^42.666666...7 unique combos.

You have just discovered an endomorphism, where two different Y's will solve the curve equation: y^2 = x^3 + 7

i.e. these two privkeys make two different but opposite Y's (because 115792089237316195423570985008687907852837564279074904382605163141518161494336 is just n-1 or, equivalently [mod n: the cyclic group 0..n, n-1...2n, etc.], it is -1).

And -1^2 = 1^2 = 1.

Now the other 3 combinations - and why only 3? see the next section - come from the X term.

Notice how the X is cubed which means it has three different roots if you consider it as a polynomial. There's obviously X, but there's also 0+Xi and 0-Xi (complex numbers). It follows the pattern [X + Yi], where the Y coord is an imaginary number.

This goes to say that if e.g. (7,0) was a valid point, then that, (0,7) and (0,n-7) would all reference similar points.

And (x,y), (y,x) and (y, n-x) would similarly reference similar points as well.

Now multiply 2*3 combos (endomorphisms) and you get a total of 6 endomorphisms: (x,y), (y,x) (y, n-x)  and (x,n-y), (n-y, x), (n-y, n-x).

It isn't something like they'd all have the same Y-point, but these points are accessible from the same X-coordinate as well. (See e.g. Roots of x^3+7 example)

There is 6-way symmetry on secp256k1 not just 2-way. Y-coordinate symmetry accounts for 3x the mirror symmetry you're talking about (which is 2).

.
.BLACKJACK ♠ FUN.
█████████
██████████████
████████████
█████████████████
████████████████▄▄
░█████████████▀░▀▀
██████████████████
░██████████████
████████████████
░██████████████
████████████
███████████████░██
██████████
CRYPTO CASINO &
SPORTS BETTING
▄▄███████▄▄
▄███████████████▄
███████████████████
█████████████████████
███████████████████████
█████████████████████████
█████████████████████████
█████████████████████████
███████████████████████
█████████████████████
███████████████████
▀███████████████▀
█████████
.
a.a
Member
**
Offline Offline

Activity: 126
Merit: 36


View Profile
October 23, 2021, 04:20:32 PM
 #8

So let's say we switch the x and y coordinates and search for the y value, we get e.g. the value 15. Then we could theoretically determine that x can be

225 = x^3 + 7
218 = x^3

3✓218 = x

X= 6.01..

( I am now to lacy to find a solution were we get an integer result for x)


Pages: [1]
  Print  
 
Jump to:  

Powered by MySQL Powered by PHP Powered by SMF 1.1.19 | SMF © 2006-2009, Simple Machines Valid XHTML 1.0! Valid CSS!