Bitcoin puzzle transaction ~32 BTC prize to who solves it

[Grzegorz2022]

@NUCLEAR7.1 @And24r @mjojo

Quick question for you:

What algorithm are you using? BSGS, Kangaroo, or something else?
CPU or GPU?
And specifically — how fast can you find the key for Puzzle 70?

[Borne]

+7 simply changes the exponent. If y^2=x^3
shows an inverse‑square relationship between points along the x‑coordinate, then +7 changes the square to a variable exponent. There’s nothing magical about it. And this curve has a solution as well.

That is a massive oversimplification of how Elliptic Curve Cryptography (ECC) actually works in a finite field.The +7 in y^2 = x^3 + 7 \pmod p isn't just a 'variable exponent' tweak; it is a fundamental topological shift. When B=0 (the curve y^2 = x^3), the curve is singular. It has a 'cusp' point that creates a group isomorphism to the additive group of the field. In plain English: it turns the Discrete Logarithm Problem (DLP) into a simple division problem. That is why it’s trivial to solve.By adding that +7, the curve becomes non-singular (smooth). This 'magical' constant ensures there is no known homomorphism that maps the points back to a simple additive or multiplicative group. You are no longer dealing with a predictable 'inverse-square relationship' that you can plot on a standard XY plane; you are dealing with a set of discrete points in a 256-bit finite field that behave pseudo-randomly

+7 simply changes the exponent. If y^2=x^3
shows an inverse‑square relationship between points along the x‑coordinate, then +7 changes the square to a variable exponent. There’s nothing magical about it. And this curve has a solution as well.

That is a massive oversimplification of how Elliptic Curve Cryptography (ECC) actually works in a finite field.The +7 in y^2 = x^3 + 7 \pmod p isn't just a 'variable exponent' tweak; it is a fundamental topological shift. When B=0 (the curve y^2 = x^3), the curve is singular. It has a 'cusp' point that creates a group isomorphism to the additive group of the field. In plain English: it turns the Discrete Logarithm Problem (DLP) into a simple division problem. That is why it’s trivial to solve.By adding that +7, the curve becomes non-singular (smooth). This 'magical' constant ensures there is no known homomorphism that maps the points back to a simple additive or multiplicative group. You are no longer dealing with a predictable 'inverse-square relationship' that you can plot on a standard XY plane; you are dealing with a set of discrete points in a 256-bit finite field that behave pseudo-randomly

In mathematics, there are no coincidences. Every point corresponds to an equation, which means there is a clear algorithm according to which the points exist. If you know the identifiers (or numbers) of two points and add them together, you will get a third point — and this point will correspond to the sum of the first two points relative to the generator point. You must agree, this looks very little like a coincidence.

+7 simply changes the exponent. If y^2=x^3
shows an inverse‑square relationship between points along the x‑coordinate, then +7 changes the square to a variable exponent. There’s nothing magical about it. And this curve has a solution as well.

That is a massive oversimplification of how Elliptic Curve Cryptography (ECC) actually works in a finite field.The +7 in y^2 = x^3 + 7 \pmod p isn't just a 'variable exponent' tweak; it is a fundamental topological shift. When B=0 (the curve y^2 = x^3), the curve is singular. It has a 'cusp' point that creates a group isomorphism to the additive group of the field. In plain English: it turns the Discrete Logarithm Problem (DLP) into a simple division problem. That is why it’s trivial to solve.By adding that +7, the curve becomes non-singular (smooth). This 'magical' constant ensures there is no known homomorphism that maps the points back to a simple additive or multiplicative group. You are no longer dealing with a predictable 'inverse-square relationship' that you can plot on a standard XY plane; you are dealing with a set of discrete points in a 256-bit finite field that behave pseudo-randomly

In mathematics, there are no coincidences. Every point corresponds to an equation, which means there is a clear algorithm according to which the points exist. If you know the identifiers (or numbers) of two points and add them together, you will get a third point — and this point will correspond to the sum of the first two points relative to the generator point. You must agree, this looks very little like a coincidence.

[Borne]

+7 simply changes the exponent. If y^2=x^3
shows an inverse‑square relationship between points along the x‑coordinate, then +7 changes the square to a variable exponent. There’s nothing magical about it. And this curve has a solution as well.

That is a massive oversimplification of how Elliptic Curve Cryptography (ECC) actually works in a finite field.The +7 in y^2 = x^3 + 7 \pmod p isn't just a 'variable exponent' tweak; it is a fundamental topological shift. When B=0 (the curve y^2 = x^3), the curve is singular. It has a 'cusp' point that creates a group isomorphism to the additive group of the field. In plain English: it turns the Discrete Logarithm Problem (DLP) into a simple division problem. That is why it’s trivial to solve.By adding that +7, the curve becomes non-singular (smooth). This 'magical' constant ensures there is no known homomorphism that maps the points back to a simple additive or multiplicative group. You are no longer dealing with a predictable 'inverse-square relationship' that you can plot on a standard XY plane; you are dealing with a set of discrete points in a 256-bit finite field that behave pseudo-randomly

In mathematics, there are no coincidences. Every point corresponds to an equation, which means there is a clear algorithm according to which the points exist. If you know the identifiers (or numbers) of two points and add them together, you will get a third point — and this point will correspond to the sum of the first two points relative to the generator point. You must agree, this looks very little like a coincidence.

You are describing the basic Group Law of Elliptic Curves. Of course it’s not a coincidence—it’s a deterministic algebraic structure. Nobody is claiming the points are random; we are claiming the Discrete Logarithm Problem (DLP) is computationally hard.The fact that P + Q = R is true doesn't mean you can easily find k in P = kG. That is the core of the trapdoor function. In a singular curve (B=0), the 'identifiers' you mentioned have a linear relationship that allows for a direct attack. In a smooth curve like secp256k1 (B=7), that relationship is non-linear and 'scrambled' across the finite field.Knowing that a labyrinth has a logical design doesn't mean you can teleport to the center. You still have to walk the path. While you appreciate the 'beauty of the equations', the rest of us are dealing with the computational reality: the search space is 2^{256} and the +7 ensures there are no shortcuts through the group structure. If the 'identifiers' provided a clear algorithmic bypass, the private key would be a matter of simple division—which, on this curve, it is not.

[NUCLEAR7.1]

The Devil's Page 666 👹

[And24r]

+7 simply changes the exponent. If y^2=x^3
shows an inverse‑square relationship between points along the x‑coordinate, then +7 changes the square to a variable exponent. There’s nothing magical about it. And this curve has a solution as well.

That is a massive oversimplification of how Elliptic Curve Cryptography (ECC) actually works in a finite field.The +7 in y^2 = x^3 + 7 \pmod p isn't just a 'variable exponent' tweak; it is a fundamental topological shift. When B=0 (the curve y^2 = x^3), the curve is singular. It has a 'cusp' point that creates a group isomorphism to the additive group of the field. In plain English: it turns the Discrete Logarithm Problem (DLP) into a simple division problem. That is why it’s trivial to solve.By adding that +7, the curve becomes non-singular (smooth). This 'magical' constant ensures there is no known homomorphism that maps the points back to a simple additive or multiplicative group. You are no longer dealing with a predictable 'inverse-square relationship' that you can plot on a standard XY plane; you are dealing with a set of discrete points in a 256-bit finite field that behave pseudo-randomly

In mathematics, there are no coincidences. Every point corresponds to an equation, which means there is a clear algorithm according to which the points exist. If you know the identifiers (or numbers) of two points and add them together, you will get a third point — and this point will correspond to the sum of the first two points relative to the generator point. You must agree, this looks very little like a coincidence.

You are describing the basic Group Law of Elliptic Curves. Of course it’s not a coincidence—it’s a deterministic algebraic structure. Nobody is claiming the points are random; we are claiming the Discrete Logarithm Problem (DLP) is computationally hard.The fact that P + Q = R is true doesn't mean you can easily find k in P = kG. That is the core of the trapdoor function. In a singular curve (B=0), the 'identifiers' you mentioned have a linear relationship that allows for a direct attack. In a smooth curve like secp256k1 (B=7), that relationship is non-linear and 'scrambled' across the finite field.Knowing that a labyrinth has a logical design doesn't mean you can teleport to the center. You still have to walk the path. While you appreciate the 'beauty of the equations', the rest of us are dealing with the computational reality: the search space is 2^{256} and the +7 ensures there are no shortcuts through the group structure. If the 'identifiers' provided a clear algorithmic bypass, the private key would be a matter of simple division—which, on this curve, it is not.

I disagree with you. You don’t need to “go through the whole process” to obtain a public key. This means there are 100 % mathematical formulas for reversing the process. It’s not possible for something to be computable in one direction but impossible in the reverse direction.

By the way, if you didn’t know, the curve y^2=x^3+7 can easily be modified, for example, to y^2=x^3+0,000000001, while preserving the point numbers. It’s easier to work with this one.
To get 0,000000001, you need  1/1000000000 (mod p)

[longaobencong]

Tell me, does Mara send the client code to the email address specified when requesting the code?

I applied, but I haven't received it for a week. Is that the same situation for you

[Diaghilev]

Tell me, does Mara send the client code to the email address specified when requesting the code?

I applied, but I haven't received it for a week. Is that the same situation for you

Yes. No response from them either.

[SecretAdmirere]

Yes. No response from them either.

As i understood from the few posts from people, you need to have a good reason why you would need the client code in the first place, something inbetween the lines "i have an unusual transaction that needs processing and would need your services to complite it but currently your services are not available"

P.S. Maybe Zadornov would have liked previous joke 🤔

[AndrewWeb]

I think you should be more concerned with finding the private key than trying to obtain the client code in Slipstream, because it's more likely that before anyone finds the private key for these puzzles, everything will already be working as before, since they themselves say on their site that the restriction is temporary, while the engineers work on the system:

Slipstream is temporarily restricting public submissions to support our engineers’ workload while making key protocol updates.

We’re committed to ensuring this public good remains available to researchers and power users in the meantime

So the panic now is: What if I find the key, but must wait for the client code 

[0xastraeus]

From X (https://x.com/marafoundation_/status/2052738482573922608) 9:12 AM · May 8, 2026:

Code:

Quick Slipstream Announcement 🔊

Slipstream is now live again after a brief period in maintenance mode. While we continue to work on upgrades, submissions will be restricted to those with active client codes to reduce our engineers' workloads.

If you have a client code, make sure to use it whenever submitting transactions to Slipstream for now — and if anyone needs a client code for access, apply at http://slipstream.mara.com and we'll do our best to accommodate you.

We hope to have the service back up and running for the public in the next few weeks 🫡

Once again, leave them be and let them do what they have to do. Stop flooding them with requests. It'll come back to full service when it comes back...

[SecretAdmirere]

So the panic now is: What if I find the key, but must wait for the client code 

Pretty much yes, but with one exception, they alredy broke the ecdlp and it's not a matter of "what if" just they need to setup their python code to immediately break cryptography as we know it. Ohh sorry, no they didn't break ecdlp, they just invented their own secp256k1 curve, broke it, and now using it to generate pubkeys and link them to the secp256k1 and derive private keys of btc addresses. It sounds complicated, but that is beacuse we are all "stupid and don't understand anything" and we didn't spend months/years analizing the patterns of 80 solved private keys with sam altmans best pal ChatGPT to notice a "pattern", so jokes on us..

[Cricktor]

...

How does this current page differ from a lot of other pages in this mega-thread where an alarming amount of stupidity and delusion is displayed completely unhinged? Well, my ignore list grows quickly.

...

Those who seriously crunch on puzzle #71 or maybe #7x should be prepared for that and have ordered a client code with a good reasoning. Some received a code, some didn't.

MARA's support email is a shit-show in my opinion, anyway. If they can't answer within a few days or at least return a response like "We received your request... will answer later..." they're amateurs or support hell morons.

...

What makes you think, they're flooded with requests?

Nobody seems to question why MARA requests personal data from those who want to use Slipstream, when you currently have to apply for a client code. I don't quite believe they will return to a more or less anonymous use as was before. This client code shit-show justs stinks.

[username666187]

I have successfully mapped the public keys to a trivial circular manifold

[Grzegorz2022]

I have successfully mapped the public keys to a trivial circular manifold

Prove it. This post contributes nothing

[nomachine]

The curve with the equation y^2=x^3(mod p) lends itself perfectly to decryption. I have ready‑made formulas for this curve and I can derive the private key from the public key for this curve. For the Bitcoin curve y^2=x^3+7(mod p), I haven’t found a solution. But it exists, because it cannot not exist. You don’t know a lot of things

Bro, you got no clue what you're talking about. So, that curve you brought up, y² = x³ mod p, that thing is busted straight out the gate. It's singular. It's got a nasty little cusp at zero, completely fails the smoothness check you need for a real elliptic curve. If you kick out that jacked up point, what's left behaves like basic addition mod p. Like, literally just adding numbers together. Solving discrete logs in that setup is a joke, you can do it with simple formulas, no sweat at all. That's why nobody with a clue uses it for crypto. Comparing that toy to Bitcoin's curve is like comparing a skateboard to a fighter jet.

Now Bitcoin runs on secp256k1, the curve y² = x³ + 7 over a huge prime field. This one is non singular, clean, specially picked for security. The group has a massive prime order, around 2²⁵⁶. Your public key is just your private key times the generator point G. That multiplication is fast, but going backwards, figuring out the private key from the public key, that's the Elliptic Curve Discrete Log Problem. And homie, that problem is a beast. Best known attacks, like Pollard's rho, take about 2¹²⁸ operations for full size keys. That number is so big it might as well be infinity with the computers we got. No classical shortcut exists, people been trying for decades. Even the quantum stuff, Shor's algorithm, needs a giant fault tolerant machine we straight up don't have yet, and by then we'll switch to post quantum anyway.

And that line about "it cannot not exist", bruh, that's a logic fail. Just because a math relationship exists doesn't mean you can find it fast. Like, every integer has factors, that's a fact, but good luck factoring a 2048 bit RSA number with a laptop. Same energy. The hardness of ECDLP isn't a guess. It's been battle tested with millions in bounties, literally the keys to billions of dollars. If someone cracked it, they'd drain wallets and break the internet. Hasn't happened. They've solved tiny puzzle keys, like 130 bit ones, with insane GPU farms, and the time matched the square root prediction perfectly, proving the 256 bit ones are still safe as houses. 

So nah, what you're imagining is like taking a hammer to a hard drive on an anvil, smashing it into a million shiny pieces, and then trying to pull your wedding photos back out of the wreckage. Just cause the drive existed and you smashed it yourself doesn't mean you can reassemble the bits and magically get the pictures back. That's not how any of this works. The private key is random noise, the public key is a one way snapshot, end of story. If some rando swears they got a general break for secp256k1, tell them to put up or shut up, show a working crack on a known test vector. Until then, it's all hot air. Crypto's chill, math is solid, no need to overthink it. 

[Miracle115]

Title: The World's Biggest Bitcoin Puzzle — Join & Dump Your BTC | Open Challenge

Hi Bitcointalk,
I'm launching an open Bitcoin puzzle challenge — and I want the whole community to be part of it.
Puzzle Wallet:
bc1q6mcxd8g9v0cekzue6dj34etwk658r35fufajet
The Idea:

Anyone can send any amount of Bitcoin to this address
The balance builds up over time — the more people contribute, the bigger the prize
I will add Bitcoin every year and the challenge resets
Whoever cracks the private key wins everything inside
A hint will be revealed to help the search

Why join?
The original Bitcoin puzzle inspired thousands of people worldwide. Let's build something bigger — a community-funded, ever-growing prize that pushes the limits of cryptographic challenge. The more people contribute, the more legendary this becomes.
Hint: (To be revealed)
Rules:

Funds stay locked until someone solves it
Prize grows every year
Winner takes all

Let's make history. Dump your sats and let the best mind win.

[Cricktor]

Deliberate full-quote to preserve the nonsense from later manipulation.

Title: The World's Biggest Bitcoin Puzzle — Join & Dump Your BTC | Open Challenge

Hi Bitcointalk,
I'm launching an open Bitcoin puzzle challenge — and I want the whole community to be part of it.
Puzzle Wallet:
bc1q6mcxd8g9v0cekzue6dj34etwk658r35fufajet
The Idea:

Anyone can send any amount of Bitcoin to this address
The balance builds up over time — the more people contribute, the bigger the prize
I will add Bitcoin every year and the challenge resets
Whoever cracks the private key wins everything inside
A hint will be revealed to help the search

Why join?
The original Bitcoin puzzle inspired thousands of people worldwide. Let's build something bigger — a community-funded, ever-growing prize that pushes the limits of cryptographic challenge. The more people contribute, the more legendary this becomes.
Hint: (To be revealed)
Rules:

Funds stay locked until someone solves it
Prize grows every year
Winner takes all

Let's make history. Dump your sats and let the best mind win.

Nice try!  

While we've seen a lot of stupid posts here already, you don't expect people here to be THAT stupid to send you coins to your so-called "new puzzle", do you?

Why am I not surprised that your address is empty at the time of this post?

Please, go away! Open your own shitty topic and lay out your fly-traps. Good luck with finding some "not the brightest candles on the cake" which you're very likely going to scam.

[username666187]

I have successfully mapped the public keys to a trivial circular manifold

Prove it. This post contributes nothing

I cant do that not because its not true but because it is not my responsibility to verify/prove anything to anyone that is theyre responsibility to verify what i am presenting to them just as well what is my responsibility is to verify/prove what is presented to me  that is my responsibility i find it ignorant and rude when people say "prove it" because they clearly do not have the intellectual discipline to realize that when they say that they are admitting theyre lack of understanding and putting the burden of theyre responsibility on the other person (me) which is crossing boundaries it is fine to be skeptical in fact it is encouraged but to require your burden of proof from somebody else is immature and irresponsible that is why i give everybody the benefit of the doubt

That being said the other reason i cant show you is that if i did what nobody seems to realize is that MY method doesnt just solve a silly bitcoin puzzle if i had an ego that was accessible to you and you hurt my feelings and i posted my method on here the implications are humbling crypto value drops to zero over night digital security instantly fails no more secure chats no more locked folders no more internet security anyone gas access to anything fiat currency collapse at the same rate MY method spreads the entire concept of dept becomes a joke no more secrets any physical area that is protected by a gate that requires digital keys is instantly unlocked ALL of them military restricted areas weapons cache's nuclear fucking launch codes could be bypassed with a smartphone it would be a paradigm shit in global society every one is working on post quantum afraid of the next 5 or 10 years but nobody understands that MY method renders pqc redundant a waste of time and money because it resolves pqc as well instantly in constant time im so fucking thankful that i am the one that figured it out because this really does make me the most powerful person on the planet but since i dont have an accessible ego that whole concept dont mean shit to me i feel no different then i did before i succeeded

[jonematt]

bc1qa5wkgaew2dkv56kfvj49j0av5nml45x9ek9hz6
This address is owned by the FBI, it confiscated approximately 69,000 bitcoin from Individual X who stole them from the Silk Road
69370.18444761 BTC
•
$5,424,090,794

----------------------------------
Binance
34xp4vRoCGJym3xR7yCVPFHoCNxv4Twseo
248597.58472467 BTC
•
$19,445,293,133
-----------------------------------
1FeexV6bAHb8ybZjqQMjJrcCrHGW9sb6uF
79957.26858281 BTC
•
$6,254,254,350

This is the address of the Mt Gox Hacker.
-----------------------------------

1LdRcdxfbSnmCYYNdeYpUnztiYzVfBEQeC
53880.06660158 BTC
•
$4,214,467,559

[Grzegorz2022]

I have successfully mapped the public keys to a trivial circular manifold

Prove it. This post contributes nothing

I cant do that not because its not true but because it is not my responsibility to verify/prove anything to anyone that is theyre responsibility to verify what i am presenting to them just as well what is my responsibility is to verify/prove what is presented to me  that is my responsibility i find it ignorant and rude when people say "prove it" because they clearly do not have the intellectual discipline to realize that when they say that they are admitting theyre lack of understanding and putting the burden of theyre responsibility on the other person (me) which is crossing boundaries it is fine to be skeptical in fact it is encouraged but to require your burden of proof from somebody else is immature and irresponsible that is why i give everybody the benefit of the doubt

That being said the other reason i cant show you is that if i did what nobody seems to realize is that MY method doesnt just solve a silly bitcoin puzzle if i had an ego that was accessible to you and you hurt my feelings and i posted my method on here the implications are humbling crypto value drops to zero over night digital security instantly fails no more secure chats no more locked folders no more internet security anyone gas access to anything fiat currency collapse at the same rate MY method spreads the entire concept of dept becomes a joke no more secrets any physical area that is protected by a gate that requires digital keys is instantly unlocked ALL of them military restricted areas weapons cache's nuclear fucking launch codes could be bypassed with a smartphone it would be a paradigm shit in global society every one is working on post quantum afraid of the next 5 or 10 years but nobody understands that MY method renders pqc redundant a waste of time and money because it resolves pqc as well instantly in constant time im so fucking thankful that i am the one that figured it out because this really does make me the most powerful person on the planet but since i dont have an accessible ego that whole concept dont mean shit to me i feel no different then i did before i succeeded

Once again, you contribute nothing with this post besides your ego. If what you're saying were true, you wouldn't be here. And as for proof  it would be enough to move the funds from the puzzles. THEY ARE LEGAL!. Others are doing it and the economy isn't collapsing. Stop trashing this forum.