can you write something more about these calculation methods

The story is unbelievable but the address exists, it has over 9000 btc and there was no withdrawal from it, if you have most of the private key, for that kind of money it is worth learning some python yourself and writing a program that will check all combinations of missing characters, it is not difficult, you can put something together from ready-made codes on the Internet. Do not give anyone any information about the key over the Internet, even if you use some software, only offline. Good luck.

1. unia europejska to trzeci świat i nikogo poważnego nie obchodzi
2. binance wprowadziło te regulacje pod koniec czerwca więc chyba za dużo się nie zmieni po prostu ludzie z trzeciego świata nie mogą kupić bezpośrednio usdt

     R: 009ac20335eb38768d2052be1dbbc3c8f6178407458e51e6b4ad22f1d91758895b
     S1: 16c91427cb20a7321029c311e757dfee67f5e3f9bad23266fad8bf8aaf5cac01
     Z1: e69364d551385880ddbb19338b633e2d6c17f1922817b3e3776851780b6aacb5
     S2: 43273c2390b15bbe7e4d38559b1d4e6c0d63aad2c586652ec423d851df065271
     Z2: 015b14bdc6f69058bfa8dcdc0e8bcd1fc87f4303804f200bfa6aadf627a8d5f6
PubKey: 036ee29b13e9d9f060d078bdcee464cb21aeafe5b5bc15206c5fa3c62f882c97c9

well, this is probably a better version of this: https://lbc.cryptoguru.org/about they apparently found something there but not much, thanks for the explanation and good luck

let's say we have a list of lost addresses/keys in some distributed database and what next? how do you want to find the key: people in the p2p network are supposed to generate keys and match? how do you want to search for keys? generating all combinations and comparing with an old, long unused public key? write something more, nobody will steal your method, I'm just curious what you came up with?

it probably won't work because there aren't enough hard drives in the world to store all those numbers in order
the only method is to calculate the key and not guess or check all the combinations
ps. how do you check that the wallet is lost?

jak odpalisz parę groszy to ci podpowiem że portfele na https://www.blockchain.com/ tak chyba wyglądały musisz mieć jeszcze hasło i adres mailowy weź sprawdź czy to tam mam podobny: ccd5afc7-729a-441d-9da9-f8f26ea03ee8

1,2 EH/s to taki duży solo miner chyba

Every time I count I get hexan, not octagon. Why ?

OP you promised to post screenshots.

My English is also through a translator. I don't know how it works, but I think you can post the photos here: https://www.talkimg.com/

Are you explaining it so as not to say anything These brothers are keys with the same value "x"? Good luck counting your keys.

Describe how you find the private key from the public key. Is it a brute force method or can you calculate it without checking all the combinations. Can you recognize the parity of the key or whether it has a positive or negative value?

The topic is good but I didn't see the light. Write something more.

if I want to subtract points t=j-d

------------------------------------------------------------------------------------------------------------------------
def subpoint (xj, yj, xd, yd):
   xt = (((yj*yj+2*yd*yj+yd*yd)*modinv (xd*xd-2*xd*xj+xj*xj,N2))-xd-xj)%N2
   yt = ((((yj+yd)*(xt-xd))*modinv(xd-xj,N2))+yd)%N2
   return (xt, yt)
-------------------------------------------------------------------------------------------------------------------------
modinv  modulo division
N2  = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F

privkey1 = 2^255
privkey2 = 2^15

easy i have table of 2^*  key

2000$ is a good price for brake bitcoin transaction .

Maybe I'm stupid, but the first example give me private key:
115792089237316195423570985008687907851824985096824206522838950949005526276248
in second example i have your right key:
1012579182250697859766212192512635217927

PS. I don't know how to calculate it

hexan/bD

MixMAx123: you right my mistake (i'm too fast reply and wrong copy paste) key : 0xd5555555555555555555555555555554463c62c03cbc85871fd9f975582d3661 give:

04 9d7b3f1498794e639b6745573b33373f2661e89fd00646045ff0b6d91c8c86ee 3ba6f0e33ef26ea7de0793e47510ed49f18854f5e0b6478fbe57a52d56a7138


any way :p3-p2 =p1 i have:
04
b7974857c88015c10573e2e1ff75c33e15f3cc976304134306e24a0f8951bf47
f44f988c488c216bb95670ad2b03fcdf04e18f042abe0f8d8671460128d784d1

sorry my mistake 04 9d7b3f1498794e639b6745573b33373f2661e89fd00646045ff0b6d91c8c86ee 3ba6f0e33ef26ea7de0793e47510ed49f18854f5e0b6478fbe57a52d56a7138