Can anyone here write a fast python  script to do the following?

We'd provide it with target public key and then it divides it by 2 but then adds n/2 to the result and divides the result by 2, however we would want to keep all the generated keys keys, nothing should be discarded.

Example of what we want the script do :

Target : p1
0291b418fd1778356ce947a5cbb46539fd29842aea168486fae91fc5317177a575

Divide by 2 = p2
02239883d26ec1fc3a513e5cc1a93975d1e4ea9850271e88a36520cd3e4eaed7d0

Adding n/2 to p2 = p3
02a714beab8015a2cdffddd772360115ddec59aa4ddf0f8fe968afd1723b576744

Dividing p3 by 2 = p4
02df418550bcb5fe3124b38da7d6ea9fe90db627c04bc048a0bba5486917b31f7e

Adding n/2 to p4 = p5
03013a363007c73a9c2fccabb7a193c35d3daaf7b40026110861488e2b2efbbc56

Now what we want is actually dividing all the keys by 2 as many times as we specify but at the same time we would want to add n/2 to each newly generated key and then divide + n/2 them as well.

Easy peasy, right? Just that if we want to drop 40 characters/bits from our target, we'd need to generate and keep 2^40 public keys. This is for educational purposes only, ( yeah right! 😉 ).

So if the puzzle creator were to spend a few satoshis from puzzle 130, you could solve the key by having the second transaction output + the existing partial WIF/key? Can you explain how do you find leakage of private key from second outputs?

I edited my previous post with the results, you could do that for any other number, but you need to know the last digit of the target, otherwise you'd be doing subtraction of 1 to f divided by 10 with your first result.

Ps, I will not study to figure out how to divide by 10m and still have a correct result, if I do, I will not share it, that'd be ECC bent and broken totally.

Anyone here knows how to divide a point by 3, 4, 5, 6, 7, 8, 9 and 10 and get a correct result?

Give me a few minutes, you will be amazed, I need to prepare the sample keys on laptop. Stay tuned.😉



Here you go


2147483648  last number  of private key in decimal is 8, that's why we need to divide 8 by 10.

035318f9b1a2697010c5ac235e9af475a8c7e5419f33d47b18d33feeb329eb99a4

0000000000000000000000000000000000000000000000000000000080000000

divide by 10
99999999999999999999999999999998d668eaf0cf91f9bd7317d25489ba2727

8/10
99999999999999999999999999999998d668eaf0cf91f9bd7317d2547ced5a5b

subtract both above, result
0260b63888a1c67e22939b795af570f36d455cdc002f3ee475517fac728aa5ce24

000000000000000000000000000000000000000000000000000000000ccccccc

Now this is really embarrassing, I'm used to look at the 7-8 ending characters  in public keys, and because I haven't slept more than 24 hours, I couldn't think.

Sorry for wasting your time, p3 is similar to the result of p1 + p2 result, they are not the same, I should have known better.  Thanks for the time.

Yeah, so you have tested it, with how many missing characters exactly? Was it 21 or 19 missing? THIS IS TRUE, you can't solve such a key "very quickly". I'd love to see how you can help! show us the magic.

P1 private key is missing, correct. Why would I post a topic about it if I had it?

P2 and p3 result in different keys when added and subtracted with each other, but when we add p1 and p2, we get p3. In other words when we have 2 and 3, addition and subtraction has only 2 results, 2 + 3 = 5, or 2 - 3 = 1,  I have 2 and 3 but there are 3 answers/results, I wanna know what is it. ( so much for clarity lol ).

Of course.
P2=

P3=

Either I'm too tired and can't think straight, or there is something wrong with the following public keys, this is the first time I see something like this.

P1=

P2=

P3=

So, p1 + p2 = p3.  Now I have p2 and p3 private keys but adding and subtracting them, +n, -n, nothing works, I even added their -n keys and got a number larger than n, but used mod n and got the correct answer.

I mean there are only 2 possible results when we have 2 numbers and only add/ subtract them, how come I get a different result?

It seems I was using ice secp256k1 files from 2022, after downloading the latest one from github I managed to get it running, but first public key I provided the script only generated 780 keys and returned error no 13, permission denied to open the txt file, my second attempt with another key generated 65535 keys and stopped, now what happened with the first run, why would it want to open the txt file?

Btw, how can I change the number of subtraction? For example which ones should I change if I wanted to subtract 50 instead of 1 and then divide?


k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))


def ters(Qx,Scalar):
     ScalarBin = str(bin(Scalar))[2:]
     le=len(ScalarBin)
     for i in range (1,le+1):
        if ScalarBin[le-i] == "0":
            Qx=ice.point_multiplication(k,Qx)
        else:
            Qx=ice.point_addition(Qx,neg1)
            Qx=ice.point_multiplication(k,Qx)
     return ice.point_to_cpub(Qx)


Should I replace all 1s with 50? Lol, I'm noob.

Your help is appreciated.👍

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2nd edit:
A hint for interested parties, 2^255, find it's associated private key and double it mod n. Cheers. 😉

Are you talking to me or advising others against me? Lol, even if OP shared both public key and the remaining of the WIF  characters, nobody could solve the key in a few thousand  years, a 256 bit private key is not something we could simply brute force.

I don't know if there is a public key, but this seems more like a sold wallet case where the buyer thinks he can get help from here to find his fortune.  No public key/ address, means there is something wrong with this case.

Can you share the public key and then the characters of your WIF which you have?

Illuminati confirmed, secretly held private group in telegram!

I am busy at the moment ( honestly I have no knowledge about signatures ) so I might solve it later.😅


I guess that's why Satoshi never touched his first million bitcoins, it's a donation to everyone!

Anyone knows what is going on with my python windows?

Aside from adding ^^^^^^^ under ECC operations on screen, everytime I fix a missing lib/module, it returns error about some line 447 or 464 in secp256k1 module, error about pkg-config which I can't install because no such file, even if I download it and add it to my folder, it wont work. Is this related to the py version? I have python 3.10, and also 3.11.  Is there a place where I could paste my script to have access to everything instead of installing and moving needed files one by one? Lol

I have some old scripts which some moron AI helped me to code back then, if I can get python scripts running without any errors I could test them.

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Sir plz, don't banned me just learn here.😅

Sorry but I'm not really being a jerk when I say what you are doing is waste of time, I wasted 2 month on WIFs, then I found out there is no shortcut unless we know the checksum which is 6 base58 characters even though the checksum is not part of the hex private key but knowing it could change everything.

Apology for calling you a spammer, you could post a one line example instead of 20 lines.

Btw, if I wanted to spend time on learning how to code, I couldn't learn the little I know about ECC.  Imagine I spent 1 hour to download and install packages and then copy them one by one to the directory of my script just to realize I am not built for such things, every person has his own sets of skills, learning how to code is not mine to master.

Thanks, I did what you said and learned a few things about python commands, after a few errors now it says module secp256k1 has no attribute 'pub2upub'  even though I correctly pasted the upub in it's place.  I went to secp256k1.py and there was no mention of pub2upub. Lol this thing burned more energy than a week of dealing with EC math.

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You can't figure out if a key is odd or even, that would make it easy to find every other digits to see if they are also odd or even.

But there is a way to make sure that the correct results are in your files and your tool had all the incorrect and correct keys.

Here is how, you know some keys can be divided by 2 a few times in a row, so when you divide a key you need to divide it by 2 at least 4 times, subtract 1 from all and do the same with all the keys.  This should occupy  less than 1 GB of storage/ RAM, and it should solve the low bit range keys in minutes, but I don't know why nobody is trying it.🤔

Hey guys, instead of wasting your eye sight on long and useless base58 WIFs which literally represent 0s in hexadecimal, let me share a little secret regarding public keys.

Here is how you can find half of your public key, it's not straight forward method but I bet many of you didn't know about it.

First we need to extract 1 and half of our public key then we can subtract our p which is 1 from it's 1.5 to get it's 0.5 half. Though we could just divide it by 2 without all this trouble, this is a hint to make you dive deeper in to this vast ocean of numbers and equations.

Target  pub:

Target priv:

Our multiplier inverse or not doesn't matter, +n will give you +n result and vice versa.

Scalar : aka n/2+1 half n +1 or 1 and half of n.

If you multiply our target pub with scalar above, you will get this :
Pub2 :

Priv :

Now if we subtract target from pub2, we will get :

Pub3, half of target :

Priv :

We didn't even use division, *chop chop and good luck diving.😅


* = hurry! Get to work.

If by any chance anyone here has the WIF checksum for #66, post it here so we could find the key in a few hours.
The only advantage if WIF is when you know the checksum, the checksum changes in different  bit ranges, I mean the base58 representation of keys with the same last 8 characters is different, so we could use wifsolver to search for the key if we had the checksum.


The world of base58 and WIF checksum is an entirely different world but it is amazing once you get to know it's secrets.😉

Speaking of spamming this thread with useless content,  please, you should consider the size of the post you are quoting!

Hey man, I wanted to try your script but error is  sympy module not found or doesn't exist. Any other way to get mod inverse working?