Topic: Bitcoin puzzle (3,350.00 BTC's ) (Read 2792 times)

Well i don't believe that this puzzle is real, but that is what i believe based in what i know about the Elliptic Curve Signatures. The OP don't want to make a message signed with that publickey, that is a strong sign that is not real. but who knows.

Regards!

Hello albert0bsd,
So, may i ask if this is a real legit puzzle that is solvable then?
or is this a fake puzzle?
and if this puzzle is solvable, what is the chances that actually someone unlock that 3350 BTC Address?
Regards!

Well i just calculate some values for k1, k2 and the privatekey but the generated private key don't generate your publickey


With any of the values k1 or k2 we can calculate the pvk with the next equation:

pvk = ( k*s - z)/r  (Mod N)

Using your rsz values and my values k1 and k2, both substitutions in equation always give me the same privatekey and well this obvious the equations assume that K2 = 2 * K1 and all the subsequent substitutions come from that supposition.

I heard from other people trying to solving this example in the testnet, that they also solved some privatekey but also they doesn't generate your publickey.

---

For those who wanna to know how or where the numbers come from, please read ECDSA: Elliptic Curve Signatures there are one equation, that we can use to solve some elements.


(1) :  s = k^-1 * (z + r * d) (mod n)

That is the equation that generate the s value. d is the privatekey from that equation we can derive the private key with all the other values s,r,z and k

(2) :  d = ksr^-1 - zr^-1 (mod n)

We have 2 TX values (s,r,z) with an unknow k, but we "know" the relationship between the two k values

(3) : k₂ = 2*K₁

Since the equation (2) solve the same private key with the 2 distinct values of r,s,z and k of our transaction we can  use it in the next way

(4) : k₁s₁r₁^-1 - z₁r₁^-1 = k₂s₂r₂^-1 - z₂r₂^-1 (mod n)

if we substitute (3) in (4) and and then we can solve for k₁

(5) : k₁ = (z₁r₁^-1 - z₂r₂^-1) / (s₁r₁^-1 - 2r₂^-1s₂) (mod n)

Once that you have the value of k1 you can use it to solve k2 using the equation (3) and use the complete set of values r,s,z and k to solve the private key in the equation (2)

---

A little code in C using gmp to show this example:


Regards!

its been 6 months.  Do you have any plans on sharing educational material with results which can be reproduced to prove the theory?

Great. I usually calculate manually, sometimes due to lack of attention it results in some errors

Almost that... although I think it's considered a puzzle

Here it is!

(((TESTNET)))

I sent two transactions to this address:muFPtSuYmYpaqdrNFtwNfm5j2fK2WMtRoz

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

My address : mmjRyw4Zgn7QQVuqTe4YLQmcn82b8aD2L6

My public key : 028ce829db535d42389defbf9ba58731f56ddb1cc7a189e5e30a85d63eb225b5d2

The k value of the second transaction is double the first: k1 * 2 = k2

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.

hi everyone..!
I am not clever in crypto, but I do understand the maths. I started reading discussion from the beginning. some people they pretended like einstein..doing lot of mathematics until I though they cracked the private key...just wasted my time reading and trying to understand this shit mathematics...finally the fund is there.... ...no body cracked till now.? why all these symbols and mathematics..? even my grand mum can write these symbols..

Doesn't seem so, funds never moved.

Riiight I see. The thread should be renamed though if there is no puzzle 

Yup for sure, if it really is a puzzle, maybe there are some more details that could help to actually solve it which they lacked to mention in the initial post here.
Puzzles are exciting man but this thread is very confusing; first everyone thought that OP created a puzzle, then it turned out they have no private keys, now it's unsure if a puzzle even exists lol.

Look, I'm not an expert on the elliptic curves. I'm just interested and learn whenever I'm able to. Your way, indeed, works now that I've rethought about k2 being replaced by k1*2. The whole problem occurs when you know nothing for both k and r, but in this case you can find d given that you have one unknown value k and two unknown values r.

(Didn't actually try your equation, but the steps seem fine by me)

I don't think that interiawp tries to find the supposedly hidden 3,350 BTC; he/she just experiments with the maths of ECC as I've noticed from other threads.

OP should at least provide the source of the puzzle, where did he find it, by whom etc.

I'm gonna be honest, I'm right now not 100% what you are trying to do and what your result tells us, but if you were able to crack the key, you should just show us by moving the funds 

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 

Why would the hash of the message matter? The fact that k1 = k2 * 2 doesn't mean that h1 = h2 * 2. Please, elaborate.

No, because you can't divide in ECC and what you're asking is, essentially, if elliptic curve division is possible. In order for someone to work out your private key from two signatures he needs to somehow calculate k. Only if k is the same in both signatures (and so does r), you can find out d.

It's the same thing as saying that there's a private key α and a private key β which is 2*α. You can't calculate any of the private keys if I just give you αG and βG.

---

You 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?

I am back!

I chose a real signature on the blockchain and forged the second signature without knowing the private key. The forged signature has passed all validations and is valid...but it's still virtually impossible to calculate with the first one! I need to improve my calculations

@interiawp If I create 2 real transactions... same private key X and with the value k1 * 2 = k2. Would you be able to calculate the value of X?

Good Try
I don't know coding , can Help me ...
why can't send private messages for other users???
I too give some article clues...
Thanks

This works with standard signature. Mine are forged signatures or you can call them blind signatures. The calculation is much more complicated! I believe it is totally solvable ... If you can get some method that can merge the calculation of a standard signature vs forged signature it would be good.

I was using the other thread  to do some calculations. Remembering that the forged signatures that I recreate do not need to sign data or prove anything ... it just needs to be forged and that they have the values r, s and h that satisfy the calculation to find a private key. I know it's confusing. I will show some calculations related to my other thread .
PRIVADE KEY = 74071287274168731384314914382498140270634658281328726941106265589917762050271

p  = 115792089237316195423570985008687907852837564279074904382605163141518161494337
z1 = 84635513758865831094131084311208775267495704821994249663954751780286420288259
r1 = 99935505760319748698811422354322418311203851828465328908708024011195996180829
s1 = 49589235156255394867995584868850296899036724345858375131186053009052960413985
z2 = 0
r2 = 115035229747891778996889965749694763606205313739267493174821202115705061416296
s2 = 38207519993275076423632821614369697864201677311262964726666122651535684123121
x  = GF(p)

x    (1+s1*z1-s2*z2)/(s2*r2-s1*r1)

x = 74071287274168731384314914382498140270634658281328726941106265589917762050271

This one has the same parameters as the 3,350 BTC puzzle same private key and the nonce of the second signature (k1 / 2 = k2) the second nonce k is half the nonce of the first signature.The difference is that here I know the k

p  = 115792089237316195423570985008687907852837564279074904382605163141518161494337
z1 = 84161583072841456669059952378962616999584763854943151345373830328904632908285
r1 = 94314914130653988673888770692000596437449719230712969855406611816122161753818      
s1 = 22494341240730831470571507988479127051360132620614139425560703058275568234720
z2 = 84635513758865831094131084311208775267495704821994249663954751780286420288259
r2 = 99935505760319748698811422354322418311203851828465328908708024011195996180829
s2 = 49589235156255394867995584868850296899036724345858375131186053009052960413985
x = GF(p)

x  (43622407236688973229510697286560312319272310986763330555167501359776293201463+s1*z1-s2*z2)/(s2*r2-s1*r1)

x = 74071287274168731384314914382498140270634658281328726941106265589917762050271

I inverted the order and replaced s1 with its own inverse  modular and did the same with s2. Then replace 1+ with k +. It is already a way to try to find some method that solves it.

Look at this one ... it's getting fun!

p  = 115792089237316195423570985008687907852837564279074904382605163141518161494337
z1 = 0
r1 = 99935505760319748698811422354322418311203851828465328908708024011195996180829      
s1 = 107074468996081319021460734830045966618222458319611877930291706090648733800102
z2 = 0
r2 = 115035229747891778996889965749694763606205313739267493174821202115705061416296      
s2 = 38207519993275076423632821614369697864201677311262964726666122651535684123121
x = GF(p)

x   (1+s1*z1-s2*z2)/(s2*r2-s1*r1)

x = 74071287274168731384314914382498140270634658281328726941106265589917762050271

Now this last one ...

p  = 115792089237316195423570985008687907852837564279074904382605163141518161494337
z1 = 84635513758865831094131084311208775267495704821994249663954751780286420288258
r1 = 99935505760319748698811422354322418311203851828465328908708024011195996180829      
s1 = 60641406722465826032271764495324651446430317390841265423474818277065347735949
z2 = 27086795414784162292297506376302057554366609881154614249233399373002336547922
r2 = 115035229747891778996889965749694763606205313739267493174821202115705061416296
s2 = 5926985887680998340381673345353182670979487968029788012609647734652828070871
x = GF(p)

x   (1+s1*z1-s2*z2)/(s2*r2-s1*r1)

x = 74071287274168731384314914382498140270634658281328726941106265589917762050271


Blind and forged signatures are not useless! it is possible to calculate them with real signatures.
I am trying to improve these methods. I still have a lot of work to do

The calculations are much more difficult than I imagined ... To try to solve this you need to have a very advanced knowledge of linear equations.
The only clue is that the nonce k of the second forged signature is half of the first signature. What remains for us is to try to find some method to explore this information. ANY INFORMATION on the private key or nonce is very valuable. I'll give you an example ... If anyone knows if the private key is even or odd or if the public key is from a private key above 57896044618658097711785492504343953926418782139537452191302581570759080747168, it is possible to factor ... and in approximately 10 minutes discover any private key.

http://www.christelbach.com/ECCalculator.aspx

This fake puzzle is not solvable.

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.....

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

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 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...

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.


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.


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 not enogh? Well then please prove otherwise.

I really wish that you was rigth.

thank you, best regards

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:


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:


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.

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

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?

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.

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.

Please do it.

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.

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
.

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.

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.


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:


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:


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

Don't waste your time with this user anymore

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.

Relay hash of signet message not in tx ?

Not possible because OP is not the owner of 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx

Good morning thanks for the puzzle I have a question ...are there s2 and h2 correct values? sorry for asking but After recalculating all the values the only one I am getting wrong  is h2

Hi. And what is a message in btx transaction

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/

It doesn't. The second R & S differs.

Yours are:
r2: b0c147a4046a541d051fa5f17906bc01aeb46fbe273ef9226610d2e0579dad88
s2: 7c9f48333e7ad71c9fc66c28efaa4032954bba147feb8f08664e3fce4619a75c

The signature from 523c3a9aa8f69af7a5b0491eeb4e49be3fb20ca5e5b0a9114e0bbdf273c1a690:
3044022070aee9c959607de56c500a2aefeaed489aab151daf42299df564106367c296e10220755799de9564d2968ceac7f8a3678d1d7b47e41fadb24ba6c694fc11eae6bcc901
040b8a0382802e12fc345e9bace8b99f6aed6b90fbfd796e8027ca9bb5f472778db863952bdb6e9 e399e34f941cab2fa6c244e65af2d15244fee2d795b3f6e222d

Didn't bother to check the hashes provided.

You are very confused.
There is no such relationship between the two ks used in the 2 signatures in the 2 transactions that were sent out of 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx, I checked. If there were the coins would have been stolen a very long time ago.

I get two variats, they are not valid.

0x81cdc964a3ed34ef41ff52b1e0962a794c0b5e8143dce2b927cb6cb2af6c8439

0xe773cf35fce567d0622203c28f67478a3361bae7e6eb4366b50e1d27eb1ed82e

I go to sleep now. Tomorrow continue work if adress was not empty.

But, Buddy, this adress has many input and output transaction, so, get R,S,Z for this adress is not so HARD, and you ca get it with 2xcoin.com

BR

@BurtW Once again I apologize!
The correct is:
(Nonce k2 of the second signature: k1 / 2 = k2)
That would be too easy! The order between s and h of the second signature is alternated ... You just need an efficient equation ... and the private key is all yours. Anyone who can calculate will own the 3,500.00 BTCs
You don't need to post formulas or equations here, just calculate and get the private key for yourself.
All the information needed to find the private key has already been exposed

If you have two valid signatures using the same private key where k' = 2k then:

From each message we can derive the z value (hash of the message) so:

First message and signature (m, r, s, z)
Second message and signature (m', r', s', z')

Therefore:  ks = z + rd_A and k's' = z' + r'd_A

Therefore:  (sk - z)/r = (s'k' - z')/r'

But in this case k' = 2k so:

(sk - z)/r = (2s'k - z')/r'

So all you have to do is solve for k.  All the other values:  s, z, r, s', z', and r' are all known.

rr'(sk - z)/r = rr'(2s'k - z')/r'

r'(sk - z) = r(2s'k - z')

r'sk - r'z = 2rs'k - rz'

r'sk - r'z +rz' = 2rs'k

k = (r'sk - r'z +rz')/2rs'

Once you know k you can simply calculate the private key, d_A = (sk - z)/r

I still do not see what your two signatures have to do with the BTC at 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx

These two things:  how to solve for the private key when you know k' = 2k and the BTC stored at 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx seem to be unrelated, right?

why your second P is differert from first P ?

Before someone appears saying that the signatures do not exist and are not real ...
https://www.blockchain.com/btc/tx/d95d58461f2d3fd476db49a741944ced4383a548937bb107d8897ea6053aeda8


And it also checks the validity of the second signature

Yes, as we have all told you before:

If there is a known mathematical relationship between the two nonce used (k₁ = k₀ + n OR k₁ = m * k₀) then you can derive the private key from the two signatures.

This is why the k value is picked at random (hopefully from a secure random number generator).

The key phrase here is "known mathematical relationship between the two nonce"

We have shown you the math for this - several times.  You should be able to do the math yourself at this point.

What does the address 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx have to do with anything?

Are you really claiming to know that there are two signatures in the block chain using the private key for 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx that were constructed with k₁ = 2 * k₀?

How would you know this?

Yup, just so people wouldn't go on and solve this straight away thinking that they would get the btc only to find out that there isn't any coins to be had in the first place.

Not that we're trying to say that you're a fraud or anything but just to make sure that you really have the funds. Most puzzles here in bitcointalk went smoothly just because the one who gave the puzzle signed a message on the address they're trying to give away.

It would be great if you are able to sign a message using this address. Use current date in the message like "Today is 03/03/2021 and I (bytcoin) own this address".

If you can't then most possibly you are wasting everyone's time here.

Edited!
I apologize to everyone!
I did almost everything "manually" is susceptible to errors
Two valid signatures with the same private key, but the nonce k of the second signature is half of the first. (Nonce k of the second signature: k1 / 2 = k2)

The second nonce k2 is half of the nonce k1 of the first signature

In decimal

Signature 1:

Nonce k1
r = 385570073629551546729230374184391439442417095537024350206131291065055332733
s = 76130260662571134678372154009160868461537800596554110575850229341703072615102
h = 42268864623620244998249895290537358742158169262361846431272029622156081203721



Signature 2:

Nonce k2 (k1 / 2 = k2)
r = 79948557280043978274449002532600374620672215928697101734828211284733327158664
s = 56368220214898939855776145385640746541620815395185097101215290078962069907292
h = 87726031595873281985439431891132599611815665254308710544703498112157039462043




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!
Great song while trying to solve this puzzle

https://i.imgur.com/THooBoC.png[/img]]https://www.youtube.com/watch?v=HWjCStB6k4o