Multiplication number to get half public key "Magic Number"

[ElDalmatino]

Hi i have read long ago, there is a number that can be used for multiply a pubkey, to get the half of it.

Maybe somebody can help me, or know the number.

[1715591969]

1715591969

[ElDalmatino]

Thx but another member send me a dm with the number it´s 57896044618658097711785492504343953926418782139537452191302581570759080747169

[ElDalmatino]

This may work if the number is even, but not if the number is odd. Or am i wrong?!

[vjudeu]

This may work if the number is even, but not if the number is odd. Or am i wrong?!

It works for every number, because it is "modulo n-value".

Code:

+---+------------------------------------------------------------------+
| a | a/2                                                              |
+---+------------------------------------------------------------------+
| 1 | 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1 |
| 2 |                                                                1 |
| 3 | 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2 |
| 4 |                                                                2 |
| 5 | 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a3 |
| 6 |                                                                3 |
| 7 | 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a4 |
| 8 |                                                                4 |
| 9 | 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a5 |
+---+------------------------------------------------------------------+

See? There is always a match. And you can check, that it works, because if you take 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1, and multiply it by two, then you will get fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364142, which modulo fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 gives you one. And the same is true in other cases.

[ElDalmatino]

Multiply by this to have your public key divided by 2, 4, 8:

Code:

Target/2 = 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1
Target/4 =
3fffffffffffffffffffffffffffffffaeabb739abd2280eeff497a3340d9051
Target/8 = 
1fffffffffffffffffffffffffffffffd755db9cd5e9140777fa4bd19a06c828 

Here is a script made by @mc,

Code:

import bitcoin



target= "03774ae7f858a9411e5ef4246b70c65aac5649980be5c17891bbec17895da008cb"



Start_Range= 2

end_Range= 33

for i in range(Start_Range, end_Range):

    Div = bitcoin.divide(target, i)
    print(f"{i}- {(str(Div))}")

The script above will divide the target by 2, 3, 4, 5, ..., 32. You can set any range you want.

Ok great but lets say i take the public key of 11 = 03774ae7f858a9411e5ef4246b70c65aac5649980be5c17891bbec17895da008cb

i get on 11 the pubkey of 1 and on 22 the "i call it ground" but what are the pubkeys between ?

It´s like i take it and do my public key and subtract 1 ! till i get a target pubkey lets say 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798,

whould be way faster with secp256k1 as ice, or do i miss something.

[ElDalmatino]

OK seems like i still don´t understand, how you change the range with only have a pubkey.
I read and read but don´t understand it, if i have a a pubkey and wanna reduce the range, it will come to a point where the dez value is odd and the whole thing go in a "never ending" story.

But i know there is a way, because you all talk about it in different posts. Maybe somebody explain it, in one easy to understand way.