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Just had problem to solve equation:
s1*k1 - r1*x1 = z1
s1*k1 - r1*x1 = z2
so r=r , s=s but diff z1 =! z2
how to take x or k?
paying $50 on bitcoin address
just PM me pleae with solve
It feels like the same thing I tried to solve months ago...https://bitcointalk.org/index.php?topic=5317743.msg56371821#msg56371821...Could you provide the values so we can try to calculate?s1*k1 - r1*x1 = z1
s1*k1 - r1*x1 = z2
so r=r , s=s but diff z1 =! z2
how to take x or k?
paying $50 on bitcoin address
just PM me pleae with solve
firstly everything is possible.
second:
"What I would really like is some method that would give me the correct y value... without having to invert or choose between y or -y (I THINK IT'S IMPOSSIBLE)" ---> if you do that means you break ecdsa. You trying solve problem that pubkey is in d<n/2 or d> n/2
third -> you can check how many times is fliped (y) (as 02 to 03 and 03 to 02) depends on subgroup during adding point . but it is not possoble, you can't check n/2 subgrups -> hundreds years.. you know ... or you are immortal:D
Technically it's a second option
It will be square root modulo prime of 8
But, the square root of 8 is not an interval. Specifically, it is around 2.82. If you mod this with a prime number, such as 11, won't you get the same 2.82? (2.82 = (11 * 0) + 2.82)Please guide me I've messed things up again.
32748224938747404814623910738487752935528512903530129802856995983256684603122^
28948022309329048855892746252171976963317496166410141009864396001977208667916=
32670510020758816978083085130507043184471273380659243275938904335757337482424
The symbol that I colored red means power. The result from your process would give a much greater number, considered infinity. It doesn't give the blue one.28948022309329048855892746252171976963317496166410141009864396001977208667916=
32670510020758816978083085130507043184471273380659243275938904335757337482424
Public key equations are always modular... you forgot to include:
p = 115792089237316195423570985008687907853269984665640564039457584007908834671663
script : x from y
Code:
## Input
y = 0x3199555CE45C38B856C9F64AC6DB27000AB6CEA10CAD76B2B6E246C9A020E707
## Field parameters
# Field modulus
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
# Cube root of 1
beta = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee
## Actual code
xcubed = (y*y - 7) % p
print ("xcubed = 0x%x" % xcubed)
x = pow(xcubed, (p + 2) / 9, p)
print ("x1 = 0x%x" % x)
print ("x2 = 0x%x" % (x * beta % p))
print ("x3 = 0x%x" % (x * beta * beta % p))
y = 0x3199555CE45C38B856C9F64AC6DB27000AB6CEA10CAD76B2B6E246C9A020E707
## Field parameters
# Field modulus
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
# Cube root of 1
beta = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee
## Actual code
xcubed = (y*y - 7) % p
print ("xcubed = 0x%x" % xcubed)
x = pow(xcubed, (p + 2) / 9, p)
print ("x1 = 0x%x" % x)
print ("x2 = 0x%x" % (x * beta % p))
print ("x3 = 0x%x" % (x * beta * beta % p))
script: y from x
Code:
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
x = 0x33709eb11e0d4439a729f21c2c443dedb727528229713f0065721ba8fa46f00e
ysquared = ((x*x*x+7) % p)
print ("ysquared= %s ", hex(ysquared) )
y = pow(ysquared, (p+1)/4, p)
print ("y1 = ",hex(y))
print ("y2 = ", hex(y * -1 % p))
y1 = y
y2 = (y * -1) % p
print(" ")
print("value x = ",hex(x))
print("value x = ",x)
print(" ")
if Integer(y1) % 2 == 0:
print("value 02 to y = ",hex(y1))
print("value 02 to y =",y1)
print(" ")
print("value 03 to y = ",hex(y2))
print("value 03 to y =",y2)
if Integer(y2) % 2 == 0:
print("value 02 to y = ",hex(y2))
print("value 02 to y =",y2)
print(" ")
print("value 03 to y = ",hex(y1))
print("value 03 to y =",y1)
#test
print (hex((x**3 + 7 - y1**2) % p) )
print (hex((x**3 + 7 - y2**2) % p))
x = 0x33709eb11e0d4439a729f21c2c443dedb727528229713f0065721ba8fa46f00e
ysquared = ((x*x*x+7) % p)
print ("ysquared= %s ", hex(ysquared) )
y = pow(ysquared, (p+1)/4, p)
print ("y1 = ",hex(y))
print ("y2 = ", hex(y * -1 % p))
y1 = y
y2 = (y * -1) % p
print(" ")
print("value x = ",hex(x))
print("value x = ",x)
print(" ")
if Integer(y1) % 2 == 0:
print("value 02 to y = ",hex(y1))
print("value 02 to y =",y1)
print(" ")
print("value 03 to y = ",hex(y2))
print("value 03 to y =",y2)
if Integer(y2) % 2 == 0:
print("value 02 to y = ",hex(y2))
print("value 02 to y =",y2)
print(" ")
print("value 03 to y = ",hex(y1))
print("value 03 to y =",y1)
#test
print (hex((x**3 + 7 - y1**2) % p) )
print (hex((x**3 + 7 - y2**2) % p))
Thanks for the script ... but I already have the first and the second I already knew
What I would really like is some method that would give me the correct y value... without having to invert or choose between y or -y (I THINK IT'S IMPOSSIBLE)
@BlackHatCoiner Or it could be like this...
p = 115792089237316195423570985008687907853269984665640564039457584007908834671663
32748224938747404814623910738487752935528512903530129802856995983256684603122^
28948022309329048855892746252171976963317496166410141009864396001977208667916=
32670510020758816978083085130507043184471273380659243275938904335757337482424
p = 115792089237316195423570985008687907853269984665640564039457584007908834671663
32748224938747404814623910738487752935528512903530129802856995983256684603122^
28948022309329048855892746252171976963317496166410141009864396001977208667916=
32670510020758816978083085130507043184471273380659243275938904335757337482424
if you guys says that is not possible how this address which out of range has managed to sent the bitcoin
d89efd5fe80d5b13fd7d40a7701a842a8ddfff4016615330485411780d1773cc: Appeared in best chain at height 682911, depth 16418, work done 1255748482679875348508874554.
in [3044022016f0c4b83a70025ed9c2245f233fd7e30241c168df4f65e7ba206a54207fa2a702204a3fe2d92516ceb60691d42e86bc3d7c209fd224855ede2a86e34daad9e2f21b01] [045601570cb47f238d2b0286db4a990fa0f3ba28d1a319f5e7cf55c2a2444da7ccc136c1dc0cbeb930e9e298043589351d81d8e0bc736ae2a1f5192e5e8b061d58]
outpoint:1a2c89a8ab17ed51daa2d2e27bb7f02708aff365918e9fc6b303a614a289ce2e:4 hash160:6fc2f6488dce92411d2498cb969b739befcc7988
out HASH160 [0d6c58795fcf0f3e9de2dfc2c6f0671fcbe85f8d] EQUAL 0.00008461 BTC
this has been sent to 32uzW2E6paeGGpbudRpRk5uEG4Yb15dFzi from 1BBwZVdBjoPxotHfrKLpHJBSy7vmc2pjex ( which is out of range address)
d89efd5fe80d5b13fd7d40a7701a842a8ddfff4016615330485411780d1773cc: Appeared in best chain at height 682911, depth 16418, work done 1255748482679875348508874554.
in [3044022016f0c4b83a70025ed9c2245f233fd7e30241c168df4f65e7ba206a54207fa2a702204a3fe2d92516ceb60691d42e86bc3d7c209fd224855ede2a86e34daad9e2f21b01] [045601570cb47f238d2b0286db4a990fa0f3ba28d1a319f5e7cf55c2a2444da7ccc136c1dc0cbeb930e9e298043589351d81d8e0bc736ae2a1f5192e5e8b061d58]
outpoint:1a2c89a8ab17ed51daa2d2e27bb7f02708aff365918e9fc6b303a614a289ce2e:4 hash160:6fc2f6488dce92411d2498cb969b739befcc7988
out HASH160 [0d6c58795fcf0f3e9de2dfc2c6f0671fcbe85f8d] EQUAL 0.00008461 BTC
this has been sent to 32uzW2E6paeGGpbudRpRk5uEG4Yb15dFzi from 1BBwZVdBjoPxotHfrKLpHJBSy7vmc2pjex ( which is out of range address)
Valid private keys in decimal:
1 to 115792089237316195423570985008687907852837564279074904382605163141518161494336
This address 1BBwZVdBjoPxotHfrKLpHJBSy7vmc2pjex is from private key 18 ... so it is valid!
This address 1FYMZEHnszCHKTBdFZ2DLrUuk3dGwYKQxh from private key 115792089237316195423570985008687907852837564279074904382605163141518161494337 ... so it is invalid!
@_Counselor Thanks for the answers... yes, I'm calculating using just x. Maybe with some more advanced equation or other methods it is possible to define exactly the value of y with just x?
Each x coordinate corresponds to two y coordinates, because of square root.
To convert between Ys, you have to calculate y2 = (y1 - secp256k1.p) * -1. One of Y is even and another is odd.
To indicate which coordinate is needed, the compressed keys starts with 02 for even Y and 03 for odd Y.
Yes...but what I would really like to understand is why private key 3 needed to invert the y value and private keys 1 and 2 didn't need to invert the y value.To convert between Ys, you have to calculate y2 = (y1 - secp256k1.p) * -1. One of Y is even and another is odd.
To indicate which coordinate is needed, the compressed keys starts with 02 for even Y and 03 for odd Y.
Why do there have y values that need to be inverted and other values that don't need to be inverted?
What method is used to define when I have to invert the y value?
I do not believe! Moderators deleted my post I wrote here... Disgraceful!
I don't see it in loyce.club archives so either it was never in this thread to begin with or it was deleted too quickly (within 5 minutes of posting).
They'll probably delete this one too... I better stay a long time out of here!
I would like to understand the math behind this...
Simply and objectively!In decimal for ease.
-------------------------------------------------------------------------------------------------------------------------------------------------------
Private key : 1
Public key :
x = 55066263022277343669578718895168534326250603453777594175500187360389116729240
55066263022277343669578718895168534326250603453777594175500187360389116729240**3 + 7 =
32748224938747404814623910738487752935528512903530129802856995983256684603122**28948022309329048855892746252171976963317496166410141009864396001977208667916=
y = 32670510020758816978083085130507043184471273380659243275938904335757337482424
OKAY
---------------------------------------------------------------------------------------------------------------------------------------------------
Private key : 2
Public key :
x = 89565891926547004231252920425935692360644145829622209833684329913297188986597
89565891926547004231252920425935692360644145829622209833684329913297188986597**3 + 7 =
57199941671890039290383617934355424684258807805258215939368959893591666662646**28948022309329048855892746252171976963317496166410141009864396001977208667916=
y = 12158399299693830322967808612713398636155367887041628176798871954788371653930
OKAY
-------------------------------------------------------------------------------------------------------------------------------------------------
Private key : 3
Public key :
x = 112711660439710606056748659173929673102114977341539408544630613555209775888121
112711660439710606056748659173929673102114977341539408544630613555209775888121**3 + 7 =
104193826873522593991639737736096919049125888873761064059040146529970392609905**28948022309329048855892746252171976963317496166410141009864396001977208667916=
y = 90209061256745311731914079131285931446821116410824268969537695047367247992253
NO OKAY
------------------------------------------------------------------------------------------------------------------------------------------------
Why doesn't private key 3 meet the criteria of the previous ones?
What math or method is used to define when you need to use -y?
Simply and objectively!In decimal for ease.
-------------------------------------------------------------------------------------------------------------------------------------------------------
Private key : 1
Public key :
x = 55066263022277343669578718895168534326250603453777594175500187360389116729240
55066263022277343669578718895168534326250603453777594175500187360389116729240**3 + 7 =
32748224938747404814623910738487752935528512903530129802856995983256684603122**28948022309329048855892746252171976963317496166410141009864396001977208667916=
y = 32670510020758816978083085130507043184471273380659243275938904335757337482424
OKAY
---------------------------------------------------------------------------------------------------------------------------------------------------
Private key : 2
Public key :
x = 89565891926547004231252920425935692360644145829622209833684329913297188986597
89565891926547004231252920425935692360644145829622209833684329913297188986597**3 + 7 =
57199941671890039290383617934355424684258807805258215939368959893591666662646**28948022309329048855892746252171976963317496166410141009864396001977208667916=
y = 12158399299693830322967808612713398636155367887041628176798871954788371653930
OKAY
-------------------------------------------------------------------------------------------------------------------------------------------------
Private key : 3
Public key :
x = 112711660439710606056748659173929673102114977341539408544630613555209775888121
112711660439710606056748659173929673102114977341539408544630613555209775888121**3 + 7 =
104193826873522593991639737736096919049125888873761064059040146529970392609905**28948022309329048855892746252171976963317496166410141009864396001977208667916=
y = 90209061256745311731914079131285931446821116410824268969537695047367247992253
NO OKAY
------------------------------------------------------------------------------------------------------------------------------------------------
Why doesn't private key 3 meet the criteria of the previous ones?
What math or method is used to define when you need to use -y?
I do have a custom ECC library in JS, it'll be straightforward to add r,s,z calculations and make a browser app that can compute these automatically, to remove manual errors.
Great. I usually calculate manually, sometimes due to lack of attention it results in some errorsYou state that you're giving away 3,350 BTC and yet, you haven't provide us a signed message. What's your excuse for not providing a signed message?
If I understood OP correctly, this is not a puzzle or challenge created by them, instead they're trying to crack a key and thinking they're very close & encouraging others to find the solution.
However wrong assumptions were made and it turned out that it's not so easy to crack a Bitcoin private key
yes
Here it is!I sent two transactions to this address:muFPtSuYmYpaqdrNFtwNfm5j2fK2WMtRoz (TESTNET)
Transaction 1: https://tbtc.bitaps.com/0faeffb396eebecdade89ef0f66aa002893c3dcb2787283f2d2e6a38edd1da1c/mmjRyw4Zgn7QQVuqTe4YLQmcn82b8aD2L6
z = 0xc8d3c14a3b190b6ea53ce4317fdd51ca1cf1a235dffbc3ce566507061f901a5a
r = 0x42c995eb98c38a8f3de7dde0f5ac63a67441a7e96b821f53931495bc6ea64cb0
s = 0x35a526e609d7022249d46d2392ff7e66b11a303d137204eb909867ae272c24b1
------------------------------------------------------------------------------------------------------------------------------------------------------------------
Transaction 2: https://tbtc.bitaps.com/4728a81966ae288b3c4d9ef781acd9cefe51b7869876abf8796f52940603014e/mmjRyw4Zgn7QQVuqTe4YLQmcn82b8aD2L6
z = 0x57b1d4f6111f1dd7b87db91931682bad285aa2587c6239f7d3beb319ff2e0834
r = 0x4f8bfa709c788b2ed59dd44a16e0c7b22ca7dece56d27144d418577a38b1d0d2
s = 0x096f65fa2a2c6c054b12981c42b0aece4ac861632a2900ced39070f0dd2e86e1
The k value of the second transaction is double the first: k1 * 2 = k2
⭐ Merited by BlackHatCoiner (1)
IT IS RESOLVED!
I would like to thank users @BlackHatCoiner and @pooya87 for trying to help me and being helpful
Yesterday when I was doing the transactions... I was doing several things at the same time and I didn't pay attention and used the RIPEMD-160 from my uncompressed address. Today with more attention... I used the RIPEMD-160 from my compressed address and worked perfectly!
https://tbtc.bitaps.com/0faeffb396eebecdade89ef0f66aa002893c3dcb2787283f2d2e6a38edd1da1c
I would like to thank users @BlackHatCoiner and @pooya87 for trying to help me and being helpful
Yesterday when I was doing the transactions... I was doing several things at the same time and I didn't pay attention and used the RIPEMD-160 from my uncompressed address. Today with more attention... I used the RIPEMD-160 from my compressed address and worked perfectly!
https://tbtc.bitaps.com/0faeffb396eebecdade89ef0f66aa002893c3dcb2787283f2d2e6a38edd1da1c
What is SAGE?
https://en.wikipedia.org/wiki/SageMath
He is asking for help with some script he made with this programming language.
COBRAS, I'm assuming your code is either not working or doesn't exist at all? I'm inclining more to the second possibility.
I didn't finish this as I have other work to do now, but I'll complete it at a later date.
Code:
# Some python code
# Create compressed public key from points
def point2compressed(x, y):
if (y % 2 == 0):
prefix = "02"
else:
prefix = "03"
hx = hex(x)[2:].zfill(64)
return prefix + hx
# create uncompressed public key from points
def point2uncompressed(x, y):
hx = hex(x)[2:].zfill(64)
hy = hex(y)[2:].zfill(64)
return "04" + hx
#NOTE: code after this line is untested and probably broken
# public key to HASH160
from hashlib import sha256
import bytearray
def hash160(pubkey):
#TODO This currently doesn't add the required prefix/suffix at the hash.
hexkey = bytearray.fromhex(pubkey)
h = sha256()
h.update(hexkey)
hash160 = h.digest()
return hash160
#pip install base58
def legacy_address(hash160):
#TODO
def nested_segwit_address(hash160):
#TODO
#pip install bech32
def native_segwit_address(hash160):
#TODO
#pip install requests
import requests
def has_balance(address):
r = requests.get("https://api.blockchair.com/bitcoin/addresses?address={}".format(address))
if r.status != 200:
throw Exception("Got Status {} from Blockchair".format(r.status))
return r.json()
# Create compressed public key from points
def point2compressed(x, y):
if (y % 2 == 0):
prefix = "02"
else:
prefix = "03"
hx = hex(x)[2:].zfill(64)
return prefix + hx
# create uncompressed public key from points
def point2uncompressed(x, y):
hx = hex(x)[2:].zfill(64)
hy = hex(y)[2:].zfill(64)
return "04" + hx
#NOTE: code after this line is untested and probably broken
# public key to HASH160
from hashlib import sha256
import bytearray
def hash160(pubkey):
#TODO This currently doesn't add the required prefix/suffix at the hash.
hexkey = bytearray.fromhex(pubkey)
h = sha256()
h.update(hexkey)
hash160 = h.digest()
return hash160
#pip install base58
def legacy_address(hash160):
#TODO
def nested_segwit_address(hash160):
#TODO
#pip install bech32
def native_segwit_address(hash160):
#TODO
#pip install requests
import requests
def has_balance(address):
r = requests.get("https://api.blockchair.com/bitcoin/addresses?address={}".format(address))
if r.status != 200:
throw Exception("Got Status {} from Blockchair".format(r.status))
return r.json()
Bro !!! Thank you very mach for your help. I was temporally basy, so I well come back to my fake base point generator and script some time late.
Regard !!!
These are invalid transactions. In the last one, you use an input from this transaction to create one output of 0.05 BTC to this address while the inputs are 0.01224182 BTC summed.
Did you use a wallet software to sign this or did it by hand?
Did you use a wallet software to sign this or did it by hand?
I did it manually and offline. Including those two that worked months ago.
I received tBTC 0.0001 and am trying to send tBTC 0.00005 to muFPtSuYmYpaqdrNFtwNfm5j2fK2WMtRoz
For about 2 years I like to create raw transactions and test some things on the TESTNET network. After months...Today I decided to create some raw transactions and do some tests as usual. But now my raw transactions are no longer accepted!
Months ago:
Raw transactions:
0100000001b9c222461de38f9fcdc121a47e27c5a08932232b00adc1e2c3fbfb55041b942201000 0006a4730440220503434f27f0d5dc23c01a5d2b2b2abb902e2816987515dafa619ffded8fc2ec6 02204c30c0b82c9afd623e140bf2d6badfa2f02414ef3ffc603e83f2f65a264ce2b801210355d77 643f9bbccde7a1f05d8fa72c0b6c53075cdc299753242320ef9c6611233ffffffff01905f010000 0000001976a914b95bfe11c932408ae44da26872b3c8cd5d61691a88ac00000000
Raw transactions:
0100000001d96c0ed0e9dda1efef19a51b68bb54cad4f384c3a943c17b77c3b77e3352711a00000 0006b483045022100dcf17de661e280dbf62e03ef1655d1baaabc301da9fc6b29a63e52e7780c11 5d022059bbf685149a0975b6305d14fa7823a9a96aa00ccf1799eb1b4336e04bb4010f01210355d 77643f9bbccde7a1f05d8fa72c0b6c53075cdc299753242320ef9c6611233ffffffff0188130000 000000001976a914b95bfe11c932408ae44da26872b3c8cd5d61691a88ac00000000
Today I've tried to send 5 transactions, but all are giving an error! I'm doing as I always did
This was my last attempt:
Did I unlearn or are things different now?
Raw:
0100000001dfee7c0aa6cb1cf8f4c9ac9339d9b6078d792a106267f93802ac75b52a6ad0de00000 0006a47304402203e408f62cf85e326dc4066636c306b85ce0b94dd94c001fbfd35da58979def3f 022075255c85e927340d4777963357cb03482566e29f93cc9f590398dfb3a89cbb970121028ce82 9db535d42389defbf9ba58731f56ddb1cc7a189e5e30a85d63eb225b5d2ffffffff018813000000 0000001976a91496a07833918317f2e0e23eea585970831da66f8988ac00000000
Months ago:
Raw transactions:
0100000001b9c222461de38f9fcdc121a47e27c5a08932232b00adc1e2c3fbfb55041b942201000 0006a4730440220503434f27f0d5dc23c01a5d2b2b2abb902e2816987515dafa619ffded8fc2ec6 02204c30c0b82c9afd623e140bf2d6badfa2f02414ef3ffc603e83f2f65a264ce2b801210355d77 643f9bbccde7a1f05d8fa72c0b6c53075cdc299753242320ef9c6611233ffffffff01905f010000 0000001976a914b95bfe11c932408ae44da26872b3c8cd5d61691a88ac00000000
Raw transactions:
0100000001d96c0ed0e9dda1efef19a51b68bb54cad4f384c3a943c17b77c3b77e3352711a00000 0006b483045022100dcf17de661e280dbf62e03ef1655d1baaabc301da9fc6b29a63e52e7780c11 5d022059bbf685149a0975b6305d14fa7823a9a96aa00ccf1799eb1b4336e04bb4010f01210355d 77643f9bbccde7a1f05d8fa72c0b6c53075cdc299753242320ef9c6611233ffffffff0188130000 000000001976a914b95bfe11c932408ae44da26872b3c8cd5d61691a88ac00000000
Today I've tried to send 5 transactions, but all are giving an error! I'm doing as I always did
This was my last attempt:
Did I unlearn or are things different now?
Raw:
0100000001dfee7c0aa6cb1cf8f4c9ac9339d9b6078d792a106267f93802ac75b52a6ad0de00000 0006a47304402203e408f62cf85e326dc4066636c306b85ce0b94dd94c001fbfd35da58979def3f 022075255c85e927340d4777963357cb03482566e29f93cc9f590398dfb3a89cbb970121028ce82 9db535d42389defbf9ba58731f56ddb1cc7a189e5e30a85d63eb225b5d2ffffffff018813000000 0000001976a91496a07833918317f2e0e23eea585970831da66f8988ac00000000