Bitcoin puzzle transaction ~32 BTC prize to who solves 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?

[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,

[digaran]

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

So are you suggesting that all the keys starting with 8, 9, a, b, c, d, e and f are all in 255 bit range and also all keys starting with 1, 2, 3, 4, 5, 6, and 7 are as well in 255 bit?



Whenever you see a private key starting with 8, 9, a, b, c, d, e, f, you can tell it's 255+ bit, even most of the keys starting with 7 are in 255+ bit. This is why I use hexadecimal over decimal, as it is easier to recognize patterns, but if you work with decimals, you can't even tell which number is greater than the other, but in hex you can count 2 by 2 to quickly determine the size and the exact range.

[mcdouglasx]

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

So are you suggesting that all the keys starting with 8, 9, a, b, c, d, e and f are all in 255 bit range and also all keys starting with 1, 2, 3, 4, 5, 6, and 7 are as well in 255 bit?



Whenever you see a private key starting with 8, 9, a, b, c, d, e, f, you can tell it's 255+ bit, even most of the keys starting with 7 are in 255+ bit. This is why I use hexadecimal over decimal, as it is easier to recognize patterns, but if you work with decimals, you can't even tell which number is greater than the other, but in hex you can count 2 by 2 to quickly determine the size and the exact range.

they are all in the 255 bit range, 256 bit is a mirror. a mathematical illusion.

therefore any key in it is in 255 bit and 256 bit simultaneously.

[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!.

Have you thought about it? You can post them here if you prefer.

[mcdouglasx]

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!.

Have you thought about it? You can post them here if you prefer.

Sorry, If I manage to solve any puzzle I will give all the information at the moment I do not have a good economic position for it.

Anyone could add x quantity to the puzzle key, generating a different public key and camouflage the puzzle with it, and by giving them information I would be solving the puzzle without knowing it.

then simply subtract that same amount from the camouflaged public key and go back to the puzzle and solve it.

[GoldTiger69]

Sorry, If I manage to solve any puzzle I will give all the information at the moment I do not have a good economic position for it.

Anyone could add x quantity to the puzzle key, generating a different public key and camouflage the puzzle with it, and by giving them information I would be solving the puzzle without knowing it.

then simply subtract that same amount from the camouflaged public key and go back to the puzzle and solve it.

Well, if that is really the problem, here is the private key for that public key:

 priv key HEX:               0x155d735095df3f8c24ee33e4dfe955da
 priv key DEC:               28399010381067550399127235982780880346

 pub key:                       0343b7d69e8372746596980d678d6cdecbffb2927916b3dbd3bc3d27bf5366b166


As you can see, I won't solve any puzzle with the info you provide; I only want to test if your script works or not.

But if you decide not to test it with me, that's OK.

Cheers.

[vsv68]

Sorry, If I manage to solve any puzzle I will give all the information at the moment I do not have a good economic position for it.

Anyone could add x quantity to the puzzle key, generating a different public key and camouflage the puzzle with it, and by giving them information I would be solving the puzzle without knowing it.

then simply subtract that same amount from the camouflaged public key and go back to the puzzle and solve it.

Well, if that is really the problem, here is the private key for that public key:

 priv key HEX:               0x155d735095df3f8c24ee33e4dfe955da
 priv key DEC:               28399010381067550399127235982780880346

 pub key:                       0343b7d69e8372746596980d678d6cdecbffb2927916b3dbd3bc3d27bf5366b166


As you can see, I won't solve any puzzle with the info you provide; I only want to test if your script works or not.

But if you decide not to test it with me, that's OK.

Cheers.

With a known private key, each person can reduce the key to any size.  Here is an example:

115 bit
pub key: 02903740144e9f301f62362813dec638747153d4f18467f32286d09b9b4412768b
priv key: 0x5735095df3f8c24ee33e4dfe955da

85 bit
pub key: 02f50518c0b808dff5bd75f10e146d33677dccbfd18370e64be0c4cc36abc9675f
priv key: 0x1f3f8c24ee33e4dfe955da

40 bit
pub key: 03a0f8f7d72f6199bb09762c8500ed22eba5253f9661f752db338bbf5eeefb1852
priv key: 0xe4dfe955da

By revealing the private key, you will not know whether the method works or not.

[GoldTiger69]

Sorry, If I manage to solve any puzzle I will give all the information at the moment I do not have a good economic position for it.

Anyone could add x quantity to the puzzle key, generating a different public key and camouflage the puzzle with it, and by giving them information I would be solving the puzzle without knowing it.

then simply subtract that same amount from the camouflaged public key and go back to the puzzle and solve it.

Well, if that is really the problem, here is the private key for that public key:

 priv key HEX:               0x155d735095df3f8c24ee33e4dfe955da
 priv key DEC:               28399010381067550399127235982780880346

 pub key:                       0343b7d69e8372746596980d678d6cdecbffb2927916b3dbd3bc3d27bf5366b166


As you can see, I won't solve any puzzle with the info you provide; I only want to test if your script works or not.

But if you decide not to test it with me, that's OK.

Cheers.

With a known private key, each person can reduce the key to any size.  Here is an example:

115 bit
pub key: 02903740144e9f301f62362813dec638747153d4f18467f32286d09b9b4412768b
priv key: 0x5735095df3f8c24ee33e4dfe955da

85 bit
pub key: 02f50518c0b808dff5bd75f10e146d33677dccbfd18370e64be0c4cc36abc9675f
priv key: 0x1f3f8c24ee33e4dfe955da

40 bit
pub key: 03a0f8f7d72f6199bb09762c8500ed22eba5253f9661f752db338bbf5eeefb1852
priv key: 0xe4dfe955da

By revealing the private key, you will not know whether the method works or not.

You are right, I didn't even think about it. On the other hand, I understand his point. I can make another pub key and see if he's willing to test his method with me or not.

[Kamoheapohea]

....
You are right, I didn't even think about it. On the other hand, I understand his point. I can make another pub key and see if he's willing to test his method with me or not.

Just sign a message with the privatekey to proof that you own the privatekey for the pubkey without exposing it.
Post the signature with the used message. Without knowledge of the message the signature can be forged.