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Topic: Bitcoin puzzle (3,350.00 BTC's ) - page 2. (Read 2792 times)

member
Activity: 211
Merit: 20
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April 16, 2021, 10:23:48 AM
#34
OP, what is the tool in the screenshots you're using to calculate the curve points? Did you make it or is it some open source java program from Github?

I too would like to get a link. It would be best if you can post publicly for community users.

http://www.christelbach.com/ECCalculator.aspx
hero member
Activity: 862
Merit: 662
April 07, 2021, 08:37:01 AM
#33
I tried to solve the first and the second signature respectively
dU = (1 - s2-1e2 + s1-1e1) * (s2-1r2 - s1-1r1)-1 (mod n)

Any one can help me please.....

This fake puzzle is not solvable.
newbie
Activity: 7
Merit: 0
April 06, 2021, 12:10:34 PM
#32
Hello! @NotATether Great to have you here in this thread... you are very welcome. I have some interesting news from the other thread. And I also have a method that almost solved this one. And more other interesting things. I will send the link of this tool that you asked for.
The other day we will talk in private

I am newbie
how to run code with sage and python
can you provide sage and python code

off topic post below

r1: 99935505760319748698811422354322418311203851828465328908708024011195996180829
                s1: 14810718830809274529170993651437030466460552688297005873719201854608653306524
                e1: 84635513758865831094131084311208775267495704821994249663954751780286420288259
                r2: 115035229747891778996889965749694763606205313739267493174821202115705061416296
                s2: 56412229366601912356674994073152925730313351483910294670205660420888695151902
                e2: 711922952377524543467576566144169816136170490747613227449590530659320692002
              s1-1: 49589235156255394867995584868850296899036724345858375131186053009052960413985
              s2-1: 75860710922369590624024015031955497020040967297713867268831531011990818769063
            s2-1e2: 24319896032458654235859288439366790171987421552616806414321622974227628294346
            s1-1e1: 33373073398809441106621025265904429856170478887328914010434069704980389675914
            s2-1r2: 102756882304321902845902604711749179835279156262963247575454606290129811589248
            s1-1r1: 109263722787838616791900575947640359553086907200677310074463510255775504782173
1 - s2-1e2 + s1-1e1: 9053177366350786870761736826537639684183057334712107596112446730752761381569
    s2-1r2 - s1-1r1: 109285248753799481477573013772796728135029813341360841883596259175872468301412
 (s2-1r2 - s1-1r1)-1: 88597492899895469960154264896435952736065060080234931949365434864574123803941
                dU: 74071287274168731384314914382498140270634658281328726941106265589917762050271


thanks in advance...


I tried to solve the first and the second signature respectively
dU = (1 - s2-1e2 + s1-1e1) * (s2-1r2 - s1-1r1)-1 (mod n)

Any one can help me please.....
jr. member
Activity: 35
Merit: 2
April 04, 2021, 01:36:02 AM
#31
OP, what is the tool in the screenshots you're using to calculate the curve points? Did you make it or is it some open source java program from Github?

I too would like to get a link. It would be best if you can post publicly for community users.
newbie
Activity: 7
Merit: 0
April 02, 2021, 04:51:33 AM
#30
Hello! @NotATether Great to have you here in this thread... you are very welcome. I have some interesting news from the other thread. And I also have a method that almost solved this one. And more other interesting things. I will send the link of this tool that you asked for.
The other day we will talk in private

I am newbie
how to run code with sage and python
can you provide sage and python code

off topic post below

r1: 99935505760319748698811422354322418311203851828465328908708024011195996180829
                s1: 14810718830809274529170993651437030466460552688297005873719201854608653306524
                e1: 84635513758865831094131084311208775267495704821994249663954751780286420288259
                r2: 115035229747891778996889965749694763606205313739267493174821202115705061416296
                s2: 56412229366601912356674994073152925730313351483910294670205660420888695151902
                e2: 711922952377524543467576566144169816136170490747613227449590530659320692002
              s1-1: 49589235156255394867995584868850296899036724345858375131186053009052960413985
              s2-1: 75860710922369590624024015031955497020040967297713867268831531011990818769063
            s2-1e2: 24319896032458654235859288439366790171987421552616806414321622974227628294346
            s1-1e1: 33373073398809441106621025265904429856170478887328914010434069704980389675914
            s2-1r2: 102756882304321902845902604711749179835279156262963247575454606290129811589248
            s1-1r1: 109263722787838616791900575947640359553086907200677310074463510255775504782173
1 - s2-1e2 + s1-1e1: 9053177366350786870761736826537639684183057334712107596112446730752761381569
    s2-1r2 - s1-1r1: 109285248753799481477573013772796728135029813341360841883596259175872468301412
 (s2-1r2 - s1-1r1)-1: 88597492899895469960154264896435952736065060080234931949365434864574123803941
                dU: 74071287274168731384314914382498140270634658281328726941106265589917762050271


thanks in advance...
member
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March 06, 2021, 07:29:05 PM
#29
I hope thread will be continued. Topic starter energy is positive for make research, mistaks are happends, yes. But crackin btc is hard task. Nonce is popular theme now, and demonstrate some makin money in previous time, new methods hope posible making money today. Exact - recover privkeys from adresses withnequel z is posible for example etc... But share research, will kill result of work, because we are in the internet and all vulnerable adresses will be drained not to who find method, but who reading shared matherial. I think is more interested on this stage is go to private telegramm chat and talk there, and invire in the chat people who realy interested this theme... Maybe moderators of this forum and other helpfull people will be join to chat...

@albert0bsd if @bitocooin(topic starter) was find a signatures with equel z his method can crack private key...
hero member
Activity: 862
Merit: 662
March 06, 2021, 01:02:21 AM
#28
Clearly the newbie created an account just for this thread

Yes im new in this forum, but im moderator in other important forum, im developer i create a tool called keyhunt in C language I know more o less what im doing

At the beginning i trust in you, check my first reply i say that the signature is valid.

I perform a signature validation with your Nonce values r,s,h of K2 and yes your Signature is valid enough to know the K2 value, that signature is not public that is why some people get confused.

But then I realize that the signature was made from the a previous one.

I need to thank you, becuase to you i learn how to make a signature that look "different" but is mathematically the same signature. That signature can be made with public data.

... in fact, mathematically it is the same signature.


Look i'm not attacking you as a person, I'm demostrating that your signature (K2) is made from the a previous one, im using real values

is the same signature because:

Quote
2*r2*s1 % n = 4d27b04d19bb13563dc45d9d768a13f53f02970024759081601c4bd59aba5f76
r1*s2 % n = 4d27b04d19bb13563dc45d9d768a13f53f02970024759081601c4bd59aba5f76

and

r2*z1 % n = 3d6319695748c424ef2a249456c3a198a6ea34933ba345a8118703ef82d28977
r1*z2 % n = 3d6319695748c424ef2a249456c3a198a6ea34933ba345a8118703ef82d28977

both sides of the equation get canceled like 0=k1*0 or k1= 0/0....

Is not enogh? Well then please prove otherwise.

I really wish that you was rigth.

thank you, best regards


legendary
Activity: 2646
Merit: 1137
All paid signature campaigns should be banned.
March 05, 2021, 05:04:32 PM
#27
bytcoin,

I appreciate your eagerness and willingness to learn about Bitcoin at a technical level and we are trying to help you understand as best we can.

What you did was take one signature and make another signature that looks different but, in fact, mathematically it is the same signature.

As stated previously:

...  is not solvable because the OP crafted the signature from a previous one.

is the same signature because:

Quote
2*r2*s1 % n = 4d27b04d19bb13563dc45d9d768a13f53f02970024759081601c4bd59aba5f76
r1*s2 % n = 4d27b04d19bb13563dc45d9d768a13f53f02970024759081601c4bd59aba5f76

and

r2*z1 % n = 3d6319695748c424ef2a249456c3a198a6ea34933ba345a8118703ef82d28977
r1*z2 % n = 3d6319695748c424ef2a249456c3a198a6ea34933ba345a8118703ef82d28977

So, do you understand now?

Also, think about this just a bit:  would Bitcoin be worth about 1 trillion dollars if it was so easy to break?  If it was this easy to break it would be broken.  If you want to see what an actual attempt to break something similar to Bitcoin looks like then take a look at this paper.  It is an attempt to figure out a way to "break" RSA - which is not exactly the same a Bitcoin but for the sake of my point it is close enough:

https://eprint.iacr.org/2021/232.pdf

Note that most experts in the field do not believe that this "breaks RSA" so don't worry about that.

What I want you to notice is the level of math it takes to even attempt to break something like an RSA, ECC, ECDSA or Bitcoin.  This is an attempt by a brilliant mathematician and crypto expert.  Face facts, you don't have the math for this, I don't have the math for this, the math involve is beyond almost everyone.

I encourage you to keep learning but, really, try not to be so melodramatic about it or you will find that everyone here that can help you will just start ignoring you altogether.

ASK QUESTIONS and LEARN.  For example I though it was interesting when you asked:

Quote
If you know the mathematical relationship between the k values in two different and valid signatures can you calculate the private key.  The answer is yes.  The math to show this is not that hard and has been show to you in the other thread (k' = k + n) and in this thread (k' = m * k).

This whole "with my very limited mathematical abilities I think that I have found a way to break Bitcoin" is not the way to go about learning about Bitcoin.  It is a good way to get ignored by the very people that are here to help you learn.

member
Activity: 211
Merit: 20
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March 05, 2021, 02:53:23 PM
#26
Hello! @NotATether Great to have you here in this thread... you are very welcome. I have some interesting news from the other thread. And I also have a method that almost solved this one. And more other interesting things. I will send the link of this tool that you asked for.
The other day we will talk in private
legendary
Activity: 1568
Merit: 6660
bitcoincleanup.com / bitmixlist.org
March 05, 2021, 10:46:31 AM
#25
OP, what is the tool in the screenshots you're using to calculate the curve points? Did you make it or is it some open source java program from Github?
legendary
Activity: 3346
Merit: 3130
March 05, 2021, 10:15:13 AM
#24
so this means none won this yet?? Maybe I am gonna get it to some university professor and let's see if he can calculate it
.

Please do it.

And please give your professor these 500 addys, they are the top 500 richest address:

https://99bitcoins.com/bitcoin-rich-list-top500/

If they can claim the bitcoins from 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx then for sure they will be able to drain all those other addys. I backup the idea about OP isn't the owner of this puzzle address until he singn a message with it. Please prove I'm wrong and sign that message.
AGD
legendary
Activity: 2070
Merit: 1164
Keeper of the Private Key
March 05, 2021, 08:41:56 AM
#23
so this means none won this yet?? Maybe I am gonna get it to some university professor and let's see if he can calculate it
.

Please do it.

There is no price and no puzzle, even though OP is desperately trying to make it look like he created a puzzle with a 3,350 BTC price.
hero member
Activity: 862
Merit: 662
March 05, 2021, 05:35:58 AM
#22
so this means none won this yet?? Maybe I am gonna get it to some university professor and let's see if he can calculate it
.

Please do it.
legendary
Activity: 2464
Merit: 3878
Hire Bitcointalk Camp. Manager @ r7promotions.com
March 05, 2021, 05:02:00 AM
#21
When did I say I have the private key? Where did I say I have the private key for the address: 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx
So, you mean to say all the bitcoin those are publicly known in addresses are at risk of being stolen? That's not how bitcoin works. Yes there are chances to have your private key to get compromise only if you lack the knowledge of protecting your private key. There is also chances to have the same private key to more than one person, but the possibility is very near to zero. The current most powerful computer is not able to solve it with the technology we have.

You are wasting time of yours and other users. Better lock the topic and move on.

I am out of this drama.
newbie
Activity: 2
Merit: 0
March 05, 2021, 01:27:21 AM
#20
so this means none won this yet?? Maybe I am gonna get it to some university professor and let's see if he can calculate it
.
legendary
Activity: 3472
Merit: 10611
March 05, 2021, 12:27:52 AM
#19
anyone who is trying to solve this puzzle.
There is no puzzles to be solved here. You asked a question last month and nobody answered, so you tried again by selecting a random address with a big balance and pretending it is "vulnerable" and you succeeded since someone did reply with a solution.

Now by continuing to insist this is a puzzle and there is a known relationship between the two ks in those two signatures (while there clearly isn't any) you are just trolling.
hero member
Activity: 862
Merit: 662
March 04, 2021, 06:10:38 PM
#18
why your second P is differert from first P ?

I think that the different P point is the previous point but halving if you see its the same n value, so the only way to get (K1 / 2 = K2) , is halving N or halving the point.

the challenge look interesting at my first try i get 0/0 so i need to check my math.


Signature 2:
The methods are practically the same as the first, the difference is that I created this signature with the nonce k being the half of the nonce k of the first signature
And to find out if the nonce k2 is really half of the nonce k1 ... A simple multiplication of points will tell you
The second signature was verified in the same way as the first



I perform a signature validation with your Nonce values r,s,h of K2 and yes your Signature is valid enough to know the K2 value, that signature is not public that is why some people get confused.

The formula to calculate a signature validatarion is in the next link: https://cryptobook.nakov.com/digital-signatures/ecdsa-sign-verify-messages#ecdsa-sign

BTW here is the relevant text:

Quote
The algorithm to verify a ECDSA signature takes as input the signed message msg + the signature {r, s} produced from the signing algorithm + the public key pubKey, corresponding to the signer's private key. The output is boolean value: valid or invalid signature. The ECDSA signature verify algorithm works as follows (with minor simplifications):
  • Calculate the message hash, with the same cryptographic hash function used during the signing: h = hash(msg)
  • Calculate the modular inverse of the signature proof: s1 =
  • Recover the random point used during the signing: R' = (h * s1) * G + (r * s1) * pubKey
  • Take from R' its x-coordinate: r' = R'.x
  • Calculate the signature validation result by comparing whether r' == r
The general idea of the signature verification is to recover the point R' using the public key and check whether it is same point R, generated randomly during the signing process.

The R' Point calculated is (b0c147a4046a541d051fa5f17906bc01aeb46fbe273ef9226610d2e0579dad88, 9ad7d585e8252cd19e6364203cfb29ac1962390b6b8122da679b53c37963316e)

Only one question, is solvable?

I get some 0 value while i was working with the ecuations i get some values like:

r1*z2 - r2*z1 = 0

Best regards


Edit is not solvable because the OP crafted the signature from a previous one.

is the same signature because:

Quote
2*r2*s1 % n = 4d27b04d19bb13563dc45d9d768a13f53f02970024759081601c4bd59aba5f76
r1*s2 % n = 4d27b04d19bb13563dc45d9d768a13f53f02970024759081601c4bd59aba5f76

and

r2*z1 % n = 3d6319695748c424ef2a249456c3a198a6ea34933ba345a8118703ef82d28977
r1*z2 % n = 3d6319695748c424ef2a249456c3a198a6ea34933ba345a8118703ef82d28977

both sides of the equation get canceled like 0=k1*0 or k1= 0/0....

Don't waste your time with this user anymore
member
Activity: 211
Merit: 20
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March 04, 2021, 11:48:38 AM
#17
When did I say I have the private key? Where did I say I have the private key for the address: 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx
Even if I had the private key, I would have no obligation to sign anything for anyone!
I am not forcing anyone to comment or participate in this thread. Please do not disturb anyone who is trying to solve this puzzle. If you think this is crazy or without logic ... there are other puzzles for you "" BITCOIN PUZZLE TRANSACTION 32 BTC "is one of the best ... Go try it.

LET'S GO TO WHAT REALLY INTERESTED!!!

Address:1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx

Information obtained from these two great sites indicates that 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx still has more than 3,350.00BTCs of balance

https://www.blockchain.com/btc/address/1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx
https://bitinfocharts.com/bitcoin/address/1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx

And they also inform that the public key of the address 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx is:
040b8a0382802e12fc345e9bace8b99f6aed6b90fbfd796e8027ca9bb5f472778db863952bdb6e9 e399e34f941cab2fa6c244e65af2d15244fee2d795b3f6e222d

Blockchain is a site known and respected by virtually the entire bitcoin community

Address, balance, public key and signature are information I obtained through Blockchain and Bitinfocharts

Signature 1:
3045022000da3994e2e3127562c5c5985905aee8adcab095c868ce42642707ffc006497d022100a 850335707cbe5956068f1244f4a96f5fd1c2df073b5fe6a87362c97693b8abe01



I used two great online HEX to DEC converters

https://www.rapidtables.com/convert/number/hex-to-decimal.html
https://www.browserling.com/tools/hex-to-dec

Converting signature and public key to decimal ...

Signature 1:
r = 385570073629551546729230374184391439442417095537024350206131291065055332733
s = 76130260662571134678372154009160868461537800596554110575850229341703072615102

Public key:
x = 5219290452886381554846642120655814292700488812830998433969278473683499906957
y = 83401511541326351499427602363312724708661860852249629118671517030652656362029

Now check if the signature is valid and belongs to this public key

h/s = 28517235349398423630439789541652069713005057721301369068264419707208655905336
r/s = 65598992890922652707901950886500083639932983506688873809736866885385723932245

G*28517235349398423630439789541652069713005057721301369068264419707208655905336

Public key * 65598992890922652707901950886500083639932983506688873809736866885385723932245

The full verification has already been done here for you...



Rx = Signature fully valid and belongs to the address 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx

I think now no one has any doubts about the first signature?

Signature 2:
The methods are practically the same as the first, the difference is that I created this signature with the nonce k being the half of the nonce k of the first signature
And to find out if the nonce k2 is really half of the nonce k1 ... A simple multiplication of points will tell you
The second signature was verified in the same way as the first



If the signature verification method remains the same and the information on this site is correct:
Blockchain
Bitinfocharts
RapidTables
Browserling
 
Here we have two signatures with the same private key, the nonce k2 being half of the nonce k of the first signature.
SECOND SIGNATURE NONCE K (K1 / 2 = K2)
English is not my language, I hope everyone has understood
This thread can help you find some method that finds k or private key of the address 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx
https://bitcointalksearch.org/topic/nonce-k-and-k-1-ecdsa-signature-5317743
Constructive criticism is welcome ... To think that this is crazy you have this right, but please do not disturb the people who are trying to help. If someone is feeling uncomfortable or just wants to troll ... there are thousands of other thread on Bitcointalk.
member
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March 04, 2021, 09:19:09 AM
#16
Address: https://bitinfocharts.com/bitcoin/address/1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx




Public key : 040b8a0382802e12fc345e9bace8b99f6aed6b90fbfd796e8027ca9bb5f472778db863952bdb6e9 e399e34f941cab2fa6c244e65af2d15244fee2d795b3f6e222d


Is not difficult! An equation solves this. I will respond to comments only after the address has been drained.
Good luck!

First and foremost, provide a signature of the following quoted message, which is verifiable at https://tools.bitcoin.com/verify-message/ or https://reinproject.org/bitcoin-signature-tool/#verify or https://tools.qz.sg/

Today is 03/03/2021 and I (bytcoin) own this address

Not possible because OP is not the owner of 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx


Relay Huh hash of signet message not in tx Huh?
AGD
legendary
Activity: 2070
Merit: 1164
Keeper of the Private Key
March 04, 2021, 09:12:32 AM
#15
Address: https://bitinfocharts.com/bitcoin/address/1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx




Public key : 040b8a0382802e12fc345e9bace8b99f6aed6b90fbfd796e8027ca9bb5f472778db863952bdb6e9 e399e34f941cab2fa6c244e65af2d15244fee2d795b3f6e222d


Is not difficult! An equation solves this. I will respond to comments only after the address has been drained.
Good luck!

First and foremost, provide a signature of the following quoted message, which is verifiable at https://tools.bitcoin.com/verify-message/ or https://reinproject.org/bitcoin-signature-tool/#verify or https://tools.qz.sg/

Today is 03/03/2021 and I (bytcoin) own this address

Not possible because OP is not the owner of 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx
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