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.