Bitcoin puzzle transaction ~32 BTC prize to who solves it

[GoldTiger69]

if I have a pub key that represents an unknown pk1
pk1=957273957373958372947

my script reduces the bits for example to

pk2=958372947
that is, remove the first 12 numbers for the example.

Among approximately 1500 keys there is one of them true that corresponds, I look for the key by brute force and once I get pk2 it is easy to deduce the private key of pk1

@mcdouglasx: Hey man, please don't be offended by my request, but can you please give me the pubkeys that your script generate out of this pubkey?  

0343b7d69e8372746596980d678d6cdecbffb2927916b3dbd3bc3d27bf5366b166

You can send me a DM with the results, I'm trying to make a test.

Thanks in advance, cheers!.

[1714876584]

1714876584

[1714876584]

1714876584

[1714876584]

1714876584

[1714876584]

1714876584

[citb0in]

you ignored my question, which proves that you cannot answer it plausibly.

There is even a way to find out the range in which the key is located, ...

Really? So tell us what the range is where #66 is lying

I apologize for not providing clarification.
You can find out the range from the public key.

There is even a way to find out the range in which the key is located, ...

Really? So tell us what the range is where #66 is lying

Currently you are just one of hundreds of messiahs and wannabe bitcoin crackers who just post nonsense. Probably on top of that an awkwardly programmed forum bot. Until you prove otherwise, you remain a "smartass". Here are some posts of you before Jan/19th/2021:

https://bitcointalk.org/index.php?action=profile;u=3342959;sa=showPosts

Just as an aside, here are your posts: other than the current one since yesterday, you last posted in January 2019. One read through your questions 2.5 years of radio silence before you speak up again here in this thread. A rogue who thinks evil

Not meening offending, just being honest. Good luck in improving your chat bot algorithm

[vhh]

There is even a way to find out the range in which the key is located, ...

Really? So tell us what the range is where #66 is lying

Not to be a smart pants, but the question is silly and the answer will follow the question :  #66 lies somewhere between 0x20000000000000000 and 0x3ffffffffffffffff .

If you can determine the range of the private key from it's public key, ECC is broken. So, I don't believe that, yet.

Exactly, to determine the range, it mean determine one single bit position and that with the correct math operations can solve any key, so it is hard to believe

Totally agree with GoldTiger69 and albert0bsd here! One single bit position will break ECC!

[batareyka]

you ignored my question, which proves that you cannot answer it plausibly.

There is even a way to find out the range in which the key is located, ...

Really? So tell us what the range is where #66 is lying

I apologize for not providing clarification.
You can find out the range from the public key.

There is even a way to find out the range in which the key is located, ...

Really? So tell us what the range is where #66 is lying

Currently you are just one of hundreds of messiahs and wannabe bitcoin crackers who just post nonsense. Probably on top of that an awkwardly programmed forum bot. Until you prove otherwise, you remain a "smartass". Here are some posts of you before Jan/19th/2021:

https://bitcointalk.org/index.php?action=profile;u=3342959;sa=showPosts

Just as an aside, here are your posts: other than the current one since yesterday, you last posted in January 2019. One read through your questions 2.5 years of radio silence before you speak up again here in this thread. A rogue who thinks evil

Not meening offending, just being honest. Good luck in improving your chat bot algorithm

I will break my word and answer once and only to you because you resorted to insults that are not justified.
In order .
1. I told you that it is impossible to assign a range to the 66th key with hash160 (I do not know such a technique). And there is no open key to number 66.
2. You want to get an algorithm, because you think that there is a grail behind it. There is no grail behind him either.
3. You resorted to searching for my messages for the entire period in order to slander me and justify your supremacy, that you are right and I am still a fraud, which will very much calm your dryness.
It is your right to do so.
But if you read my messages, you noticed that I asked there if I know two different private keys that form the same public key and everything is the same up to the address, whether it is possible to damage the integrity of the curve. I received the answer that no. But everyone immediately, as now, was bombarded with questions, how did you find it, tell us and things like that, when I refused, everyone started calling me a fraud, just like now. Surely the human essence, if it does not get what it wants, then it resorts to insults.
Then I had to ask the moderator to delete the message because there was a very large amount of slander and insults. Now it turns out that this is exactly what should be done.
4. As for why I didn't write for so long, everything is simple.
After the slander, I took a break and had lunch on my soul.
and if you notice again, my last message was January 19, 2022, a month later, the war began.
So I live in a country where war has been going on for more than 500 days.
Somehow it was not as written between shelling and bombs.
Now I have calmed down a bit, so I decided to look at the forum and write, and everyone has already seen what happened.
On the one hand, you are being bombarded, on the other hand, they are insulting you. An ideal picture of the universe.

PS
There will be no further answers. Anyone who wants can write anything, even insults in my direction, if it will make him feel better and he will please his pride, please.
If the moderator deems it necessary, he can delete my posts so that they do not give rise to further insults in my direction and deviation of the forum topic from the main goal.

Thank you to everyone who understood my answer.
All my posts written above in no way intended to offend or humiliate any of the forum members or visitors.
Thanks for the peace everyone.

[btcmitko]

There is even a way to find out the range in which the key is located, ...

Really? So tell us what the range is where #66 is lying

Not to be a smart pants, but the question is silly and the answer will follow the question :  #66 lies somewhere between 0x20000000000000000 and 0x3ffffffffffffffff .

If you can determine the range of the private key from it's public key, ECC is broken. So, I don't believe that, yet.

Exactly, to determine the range, it mean determine one single bit position and that with the correct math operations can solve any key, so it is hard to believe

Totally agree with GoldTiger69 and albert0bsd here! One single bit position will break ECC!

Here is a pubkey 03995B8A5AD00A205BFB9837014E2978273B816CFB29CA9644AB6D44977C006C64

The first bit of the private key is 1

I will even give you range 2^144-2^145

Come on break the ECC with that bit.....

[vhh]

There is even a way to find out the range in which the key is located, ...

Really? So tell us what the range is where #66 is lying

Not to be a smart pants, but the question is silly and the answer will follow the question :  #66 lies somewhere between 0x20000000000000000 and 0x3ffffffffffffffff .

If you can determine the range of the private key from it's public key, ECC is broken. So, I don't believe that, yet.

Exactly, to determine the range, it mean determine one single bit position and that with the correct math operations can solve any key, so it is hard to believe

Totally agree with GoldTiger69 and albert0bsd here! One single bit position will break ECC!

Here is a pubkey 03995B8A5AD00A205BFB9837014E2978273B816CFB29CA9644AB6D44977C006C64

The first bit of the private key is 1

I will even give you range 2^144-2^145

Come on break the ECC with that bit.....

Sure : follow the same logic / algorithm that you applied when you discovered that the first bit is 1.
Nobody poses such an algorithm as we speak, modular math on prime fields forbids it.

[vhh]

guys, we don't talk about knowing the first bit, but finding it using an algorithm. It's pointless to spit out a public key when you know the private key that generates it. Take the pub key of puzzle #120 and post the first bit (discard the 0s MSB , of course). If you manage to find that bit you can extend the algo to find out all the other remaining bits.

[vsv68]

@batareyka
If you really have a way to find out the range in which the private key is from the public key, then contact me.  I will show you how using this method of yours you can get a private key from any public.

[GoldTiger69]

@batareyka
If you really have a way to find out the range in which the private key is from the public key, then contact me.  I will show you how using this method of yours you can get a private key from any public.

I'm pretty sure that someone who can manage to obtain the range of the private key out of the public key can find a method to get such private key, don't you think so?

[vsv68]

@batareyka
If you really have a way to find out the range in which the private key is from the public key, then contact me.  I will show you how using this method of yours you can get a private key from any public.

I'm pretty sure that someone who can manage to obtain the range of the private key out of the public key can find a method to get such private key, don't you think so?

Agree.  But suddenly he really found such a way but does not understand how to apply it?

[GoldTiger69]

@batareyka
If you really have a way to find out the range in which the private key is from the public key, then contact me.  I will show you how using this method of yours you can get a private key from any public.

I'm pretty sure that someone who can manage to obtain the range of the private key out of the public key can find a method to get such private key, don't you think so?

Agree.  But suddenly he really found such a way but does not understand how to apply it?

Maybe he just think he found it. As he is not willing to explain what he really found, who knows?

[vsv68]

@batareyka
If you really have a way to find out the range in which the private key is from the public key, then contact me.  I will show you how using this method of yours you can get a private key from any public.

I'm pretty sure that someone who can manage to obtain the range of the private key out of the public key can find a method to get such private key, don't you think so?

Agree.  But suddenly he really found such a way but does not understand how to apply it?

Maybe he just think he found it. As he is not willing to explain what he really found, who knows?

He made a pretty serious claim, but provided no evidence.  As long as he is silent, he exposes himself in a bad light, but at the same time he is offended by others ...

[BTC_Backdoor]

If you look at the messages and ECC knowledge level of batareyka, you can easily figure that he is neither proficient in maths not in programming. For those who have been hammering around EC they can easily point out that EC has countless ways to surprise you when you think you got it but the very next level discovery shuts you down. I am convinced he has just found one of those surprises and didn't tried to reconfirm it. Whatever subtraction, addition, multiplication & division you do, Modular arithmetic will shut you down and at the end you are nowhere. Finally you don't get offended with Questions, the world we see today is a result of questioning.

[digaran]

Guys I'm still waiting for my answer, since I revealed the exact range of my public key, it is no use if anyone can tell me the range.

03cefd304a9a8da40666e97b2d9650e31a71a9a9fcd29a24724d31a2fabbc8ea0e

I just want you to drop the second digit from my private key having just my public key above.

Mathematical problems are solved in decimals, if you work in hex or binary you will only confuse your research, once you get results you apply the conversions, if you work in hex or binary you overlook important details such as the sum of the Y coordinates of a mirrored pubkey (negative) Y+(-Y) always equals

or why? these pk share the same X coordinates.

57896044618658097711785492504343953926418782139537452191302581570759080747172

57896044618658097711785492504343953926418782139537452191302581570759080747165

I'm a bit lost after reading your reply, what is this exactly
7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a4

And this one?
7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b209d

Why did you post 2 random keys with only 7 keys difference between them?
I'm asking this because my private key has 7 in it, can you just tell me where did you get the 2 private keys above?


To the rest of you guys, don't fight and argue, there are hundreds if bitcoins waiting for you to collect, you can show your skills by collecting them.😉

[sssergy2705]

Guys I'm still waiting for my answer, since I revealed the exact range of my public key, it is no use if anyone can tell me the range.

03cefd304a9a8da40666e97b2d9650e31a71a9a9fcd29a24724d31a2fabbc8ea0e

I just want you to drop the second digit from my private key having just my public key above.

Mathematical problems are solved in decimals, if you work in hex or binary you will only confuse your research, once you get results you apply the conversions, if you work in hex or binary you overlook important details such as the sum of the Y coordinates of a mirrored pubkey (negative) Y+(-Y) always equals

or why? these pk share the same X coordinates.

57896044618658097711785492504343953926418782139537452191302581570759080747172

57896044618658097711785492504343953926418782139537452191302581570759080747165

I'm a bit lost after reading your reply, what is this exactly
7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a4

And this one?
7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b209d

Why did you post 2 random keys with only 7 keys difference between them?
I'm asking this because my private key has 7 in it, can you just tell me where did you get the 2 private keys above?


To the rest of you guys, don't fight and argue, there are hundreds if bitcoins waiting for you to collect, you can show your skills by collecting them.😉

As I see it, it's just 7 divided in half.
https://privatekeys.pw/keys/bitcoin/1286578769303513282484122055652087865031528491989721159806724034905757349938#7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0

[dextronomous]

Guys I'm still waiting for my answer, since I revealed the exact range of my public key, it is no use if anyone can tell me the range.

03cefd304a9a8da40666e97b2d9650e31a71a9a9fcd29a24724d31a2fabbc8ea0e

I just want you to drop the second digit from my private key having just my public key above.

Mathematical problems are solved in decimals, if you work in hex or binary you will only confuse your research, once you get results you apply the conversions, if you work in hex or binary you overlook important details such as the sum of the Y coordinates of a mirrored pubkey (negative) Y+(-Y) always equals

or why? these pk share the same X coordinates.

57896044618658097711785492504343953926418782139537452191302581570759080747172

57896044618658097711785492504343953926418782139537452191302581570759080747165

I'm a bit lost after reading your reply, what is this exactly
7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a4

And this one?
7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b209d

Why did you post 2 random keys with only 7 keys difference between them?
I'm asking this because my private key has 7 in it, can you just tell me where did you get the 2 private keys above?


To the rest of you guys, don't fight and argue, there are hundreds if bitcoins waiting for you to collect, you can show your skills by collecting them.😉

As I see it, it's just 7 divided in half.
https://privatekeys.pw/keys/bitcoin/1286578769303513282484122055652087865031528491989721159806724034905757349938#7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0

    (dec)HALFINV:  57896044618658097711785492504343953926418782139537452191302581570759080747169 Length Bits =  255
    (dec)HALF   :    57896044618658097711785492504343953926418782139537452191302581570759080747168 Length Bits =  255

    Privatekey (dec)k3:  57896044618658097711785492504343953926418782139537452191302581570759080747168 Length Bits =  255
    Privatekey (hex)k3:  7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0

    k3 X:  86918276961810349294276103416548851884759982251107
    k3 Y:  28597260016173315074988046521176122746119865902901063272803125467328307387891
    Binary k3:  1111111111111111111111111111111111111111111111111111111111111111111111111111111 1111111111111111111111111111111111111111111111110101110101010111011011100111001 1010101111010010001010000000111011101111111101001001011110100011001101000000110 110010000010100000



Privatekey (dec)k3h:  57896044618658097711785492504343953926418782139537452191302581570759080747169 Length Bits =  255
    Privatekey (hex)k3h:  7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1

    k3h X:  86918276961810349294276103416548851884759982251107
    k3 Y:  87194829221142880348582938487511785107150118762739500766654458540580527283772
    Binary k3h:  1111111111111111111111111111111111111111111111111111111111111111111111111111111 1111111111111111111111111111111111111111111111110101110101010111011011100111001 1010101111010010001010000000111011101111111101001001011110100011001101000000110 110010000010100001

# Curve-secp256k1

modular elliptic curve

Total of all the wallets n is the last number.
n= 115792089237316195423570985008687907852837564279074904382605163141518161494337 (In Dec)

n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 (In HEX)

Half way of n
n//2 = 57896044618658097711785492504343953926418782139537452191302581570759080747169

57896044618658097711785492504343953926418782139537452191302581570759080747169 Lenght Bits =  255

So we know half way is 255 Bit so 50% of wallets in 255-256 Bits and the other 50% in 1-255 Bits.

What if we could divide by bits the range all the way to 1 bit? All keys are the modular inverse of the first key start point.

Thanks to Quite the Contrary for teaching me how to do this https://youtu.be/Vlqy1zB-QkE  ecdsa secp256k1 algorithm explained.
 

[digaran]

What if we could divide by bits the range all the way to 1 bit? All keys are the modular inverse of the first key start point.

Thanks to Quite the Contrary for teaching me how to do this https://youtu.be/Vlqy1zB-QkE  ecdsa secp256k1 algorithm explained.
 

How do you propose we divide by bits down to 1? And who taught you what exactly, the youtube you meant?

As I see it, it's just 7 divided in half

He just subtracted 7 from N and divided the result by 2, and as I see it, it's just 15 divided in half, because dividing 7 in half would give you 3.5 where you could see 3 as the first character and that .5 as a huge number, here is the example :
3fffffffffffffffffffffffffffffffaeabb739abd2280eeff497a3340d9052

As you see, the .5 which is a fraction has turned to something different, this is the beauty of big numbers combined with elliptic curves.😉

[sssergy2705]

What if we could divide by bits the range all the way to 1 bit? All keys are the modular inverse of the first key start point.

Thanks to Quite the Contrary for teaching me how to do this https://youtu.be/Vlqy1zB-QkE  ecdsa secp256k1 algorithm explained.
 

How do you propose we divide by bits down to 1? And who taught you what exactly, the youtube you meant?

As I see it, it's just 7 divided in half

He just subtracted 7 from N and divided the result by 2, and as I see it, it's just 15 divided in half, because dividing 7 in half would give you 3.5 where you could see 3 as the first character and that .5 as a huge number, here is the example :
3fffffffffffffffffffffffffffffffaeabb739abd2280eeff497a3340d9052

As you see, the .5 which is a fraction has turned to something different, this is the beauty of big numbers combined with elliptic curves.😉

0x7/2
./keymath 025cbdf0646e5db4eaa398f365f2ea7a0e3d419b7e0330e39ce92bddedcac4f9bc / 2
Result: 03592152c398d6c719636a03a6dad64246a5a6814aa62c156b0ce5332f6759b031
We get the result divided in half in the middle of the key range of the bitcoin curve.
https://privatekeys.pw/key/7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a4#public

[mcdouglasx]

Guys I'm still waiting for my answer, since I revealed the exact range of my public key, it is no use if anyone can tell me the range.

03cefd304a9a8da40666e97b2d9650e31a71a9a9fcd29a24724d31a2fabbc8ea0e

I just want you to drop the second digit from my private key having just my public key above.

Mathematical problems are solved in decimals, if you work in hex or binary you will only confuse your research, once you get results you apply the conversions, if you work in hex or binary you overlook important details such as the sum of the Y coordinates of a mirrored pubkey (negative) Y+(-Y) always equals

or why? these pk share the same X coordinates.

57896044618658097711785492504343953926418782139537452191302581570759080747172

57896044618658097711785492504343953926418782139537452191302581570759080747165

I'm a bit lost after reading your reply, what is this exactly
7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a4

And this one?
7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b209d

Why did you post 2 random keys with only 7 keys difference between them?
I'm asking this because my private key has 7 in it, can you just tell me where did you get the 2 private keys above?


To the rest of you guys, don't fight and argue, there are hundreds if bitcoins waiting for you to collect, you can show your skills by collecting them.😉

As I see it, it's just 7 divided in half.
https://privatekeys.pw/keys/bitcoin/1286578769303513282484122055652087865031528491989721159806724034905757349938#7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0

    (dec)HALFINV:  57896044618658097711785492504343953926418782139537452191302581570759080747169 Length Bits =  255
    (dec)HALF   :    57896044618658097711785492504343953926418782139537452191302581570759080747168 Length Bits =  255

    Privatekey (dec)k3:  57896044618658097711785492504343953926418782139537452191302581570759080747168 Length Bits =  255
    Privatekey (hex)k3:  7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0

    k3 X:  86918276961810349294276103416548851884759982251107
    k3 Y:  28597260016173315074988046521176122746119865902901063272803125467328307387891
    Binary k3:  1111111111111111111111111111111111111111111111111111111111111111111111111111111 1111111111111111111111111111111111111111111111110101110101010111011011100111001 1010101111010010001010000000111011101111111101001001011110100011001101000000110 110010000010100000



Privatekey (dec)k3h:  57896044618658097711785492504343953926418782139537452191302581570759080747169 Length Bits =  255
    Privatekey (hex)k3h:  7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1

    k3h X:  86918276961810349294276103416548851884759982251107
    k3 Y:  87194829221142880348582938487511785107150118762739500766654458540580527283772
    Binary k3h:  1111111111111111111111111111111111111111111111111111111111111111111111111111111 1111111111111111111111111111111111111111111111110101110101010111011011100111001 1010101111010010001010000000111011101111111101001001011110100011001101000000110 110010000010100001

# Curve-secp256k1

modular elliptic curve

Total of all the wallets n is the last number.
n= 115792089237316195423570985008687907852837564279074904382605163141518161494337 (In Dec)

n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 (In HEX)

Half way of n
n//2 = 57896044618658097711785492504343953926418782139537452191302581570759080747169

57896044618658097711785492504343953926418782139537452191302581570759080747169 Lenght Bits =  255

So we know half way is 255 Bit so 50% of wallets in 255-256 Bits and the other 50% in 1-255 Bits.

What if we could divide by bits the range all the way to 1 bit? All keys are the modular inverse of the first key start point.

Thanks to Quite the Contrary for teaching me how to do this https://youtu.be/Vlqy1zB-QkE  ecdsa secp256k1 algorithm explained.
 

All keys are in 255 Bit , 256 bit is a mirrow

[GR Sasa]

For the love of Almighty God, for the love of everything, Please guys help me; someone help this newbie compile the ecctools made by Alberto for windows,

https://github.com/albertobsd/ecctools

I tried compiling myself but it fails. I don't know, I'm not programmer.

Many thanks,