Quote from: dextronomous on Today at 01:01:51 PM

Quote from: sssergy2705 on Today at 09:51:27 AM

Quote from: digaran on Today at 08:53:02 AM

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.

Quote from: mcdouglasx on July 12, 2023, 01:47:37 PM

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.
Shocked

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