Reduced what exactly, addresses? You don't have public key, so reducing whatever you think you reduced is not gonna work. But hey, I wish you are up to something and really hope to find the keys you are looking for.

There is no such a thing as multiplying 2 public keys to have a valid result, you can't multiply public key 3 with 2 to have the public key of private key 6.

So why are you asking for something that doesn't exist?

Cool project, keeping an eye on this one!
Btw what project? 🤣 first show us what you got, then we'll talk about support, in case you haven't noticed, people are here to find money not give it away.😉

So this is the reason when I multiply G by itself I can't figure out what is the result! I wonder if there is a way to find the result private key?🤔

Edit, Ok, I admit that I didn't spend enough time to figure it out, after posting this I went out to find out G * G is what?

Here it is  public key :

Private key :

I hope by multiplying point by point, you meant public keys by public keys? if I misunderstood and this is off topic, apology.

I just read WP's page, there was only 1 example, since Alberto and WP  share the same topic, I had to ask it here, you could have said that when I asked you via PM. Thanks though, let the never ending division begin! 😉

Is there any example command to use modmath, keydivision etc for windows? Appreciate a reply.

Ok, how do you know the result is -800? And whatever the result, it's not invalid, but what if the key is 1890? You'd be searching from 700 to 800 in hopes of finding you imaginary -800, while the actual result is 110, far away from your searching range. I guess you haven't figured out a way to  validate whether your target is greater than 1500 or not, once you figure that out, you can base your future calculations on the fact that your target is greater or smaller than 1500.

E.g. if the key is 1650, multiply by 3 to get 4950, divide by 2 to get 2475, subtract your target from 2475 to get 825, which is half of your target.
Or just subtract 1 divide by 2, to name a few methods.



So the expert mathematician would use Y and then adds 1? I wonder about the logic in doing that, he might be crazy if he thinks that he can use a different algorithm to find secp256k1 keys faster. Are you sure this "good mathematician" knows about mod n in secp256k1 curve?
Anyways, when searching for unkown keys, doesn't matter if the keys are greater than n or not, because you don't have the pk to check, though if we use  standard implementation then we can be sure all keys are valid since they are all calculated mod n, whatever you do, public keys are always mod p, otherwise you get invalid public keys.

When we are talking about bitcoin's curve, you can NOT use different values expecting the results to work on bitcoin keys.

When I say misinformation, now I know you are misinformed, this key :
is an invalid key, but mod n it is the same as G or 0x1. You see, whatever you add to n, it's like adding to zero, if you add 5 to n, what you get is just 5. So no, they are not all the same, if you convert the key above to WIF no wallet will accept it because it has a wrong format. All pk are mod n.

How exactly did you come up with those numbers? Since when 3=7?
If you don't know how it works don't spread misinformation!

This is -n private key of 0x7,
And this is +n private key of 0x7,
Now where is 3 here?
If you divide 7 by 2, you will have -n of 0x3, which you will obtain by subtracting 4 from N =

One useful trick is dividing an odd key without having any remainder, that is dividing by 2 then adding n/2, or just subtract 1 from pk then divide by 2.

The only advantage we have regarding the range of keys is that we know the exact range for each one of them yet we can't do anything about it.

Do you have any tricks to reduce the range bit by bit? Well there is one, subtraction is the key, but you'd need to know what to subtract from your key. 😉

What is your own benchmark? I can't see it, impress us with your speed rate first, otherwise it's not better than bitcrack, while bitcrack is open source and yours is not.!

Are you searching for #66 in 63 bit range? And why only for #66, are you incapable of developing something for higher ranges?

Hello, is there a way to use available calculators on secretscan offline? I have tried but it won't work, have you released their code somewhere or we should just use the site online?
There are some useful tools which are really amazing and easy to use, wish we could use them offline too.

Here is a problem I'm facing, it seems really easy to figure out, but I can't. Maybe you could give me a solution considering you are really good with math.

I have this key :
When I subtract it from this key :
I get this result :  *first result

Now here is another key :

If we subtract the above from this one :
We will reach this key :  which is 1/8 of *first result.

Do you think there is a way to find any of the public keys above?

Nvidia GPU, also not open source, modified version of 2 different tools.
https://bitcointalk.org/index.php?topic=5328080.msg56699222#msg56699222

Bitcrack :
https://bitcointalk.org/index.php?topic=4453897.0

I'd suggest to dive into the ocean of public keys, learn how to add/subtract/divide, your chance of success is better than brute forcing for keys.

Did you just use good cop, bad cop method on him? Lol.
If he wanted to share the code, he'd have done it at the first place and nobody would have attacked him.

In case some of you haven't noticed, around these woods we paint toads and sell them as porsche!

Personally I have nothing against him, I know enough not to run the tool, even if I do I have nothing of value on my system.
Besides, his tool, your tool, my tool, none of them really help unless you have 100+ GPUs, then having that many GPUs, you wouldn't go for #66, you'd use them to search for #130. So if we are talking about "luck" searching a site like keys.lol should do the trick, if it's up to your "luck" clicking on page numbers is enough or buying a lottery ticket could also give a better result. Being "lucky" is not exclusive to finding a puzzle key, So your and his argument is not reasonable.

By 2140? Lol, it has nothing to do with "network complexity" whatever that means.

Simple and pure mathematics is what keeps the coins safe, in order to make it harder for quantum computers we just need more complex math/equations.

Enigma in WW2, was the ultimate crypto/math problem, but the first computer invented managed to break the encryption, now it's the other way around, we need to invent an equation so the new computer generation can't break it.

Satoshi just chose one of the strongest curves at that time, even he knew 20 years later people will have to change the key to their safe.😉

Why don't you share this different approach in details with the rest of us? If it was working, you wouldn't be here talking, instead you'd have rented hundreds of CPUs to find #66.

I tried inversion and it doesn't work, if you divide +n odd key by 5 and then add n/5 to the result, you won't get near the 1/5 of your +n key.
I always use inversion and I'd never divide by anything other than n/2.


I might have, but I don't recall, as I said I'm like a student learning here.😉

The result is not random at all, did you even read my first post and other posts?

Let me give you a solution in order to successfully divide an odd key by 2, I have tried 4, 5, 6 etc, I couldn't get any meaningful result, I'm still learning.

Now the example, you could either add 1 or subtract 1 from your odd private (public key) key and then divide by 2 or you could simply divide the odd key by 2 add n/2 to the result, you will get the same result.

If your key is divisible  by 2, 4, 8 etc but the last digit is 1, you could reach key 0x3 by dividing by 2 adding n/2 and dividing the result again by 2 adding n/2 till you reach the lowest point.

Demonstration :

Private key :
Public key :
Divided by 2, result:
Adding n/2 to above
N/2 =

Result, private key :
Result, public key :

And if you keep dividing you will eventually reach 0x3, dividing 3 by 2 will give you 0x2 then 1, and dividing 1 by 2 will give you half, aka n/2 and take a look at 0.5 = n/2, it is a number around 2^255.

Now to simplify it, imagine you have 23 as your odd key, dividing by 2 adding n/2 will give you 12, it's like adding 1 to 23 = 24/2 = 12. You have 2 options.

Furthermore, there are ways to accurately divide your key by 2 without any division, but that rock is really big and I'm afraid the glass will shatter, you should start with small rocks, learn from Musk throwing a big rock at E-truck's window claiming it won't shatter. Lol

Speed? No benchmark? We already have GPU enabled tools which are 20 to 30 times faster than CPU versions, I don't know what you are promising, but definitely your tool is not faster than others whatsoever, besides searching with CPU is useless.