Public Key Division - Number Theore

[GR Sasa]

Thanks for nice answers everyone,

Yes, though I thought i was smart and decided to try to keep dividing public key by 2  to the lowest point reaching base point but I Failed Everytime haha.

I think the only way is to guess ODD VS EVEN but that's impossible since private keys are random numbers.  

I think it's like guessing 0 or 1 256 times, since there are 5 odd numbers and 5 even numbers in each new value. So the complexity stays the same as before.

Unless of course someone finds a way to know Odd private keys from EVEN, then he can divide by exact reaching his targeting key.

I like your examples guys!

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[digaran]

Just use inversion:

Code:

   n=fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
 n/2=7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1
-n/2=7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0
 n/4=bfffffffffffffffffffffffffffffff0c0325ad0376782ccfddc6e99c28b0f1
 n/5=66666666666666666666666666666665e445f1f5dfb6a67e4cba8c385348e6e7
 n/6=d5555555555555555555555555555554463c62c03cbc85871fd9f975582d3661

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.

By the way, last time you messed up 1/2 with -1/2

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

[garlonicon]

I tried inversion and it doesn't work

It has to work. Let's test it on some key, where none of us knows the private key:

Code:

03 678AFDB0FE5548271967F1A67130B7105CD6A828E03909A67962E0EA1F61DEB6   Satoshi
02 2959350037EACA04CF26538AECA4402432BC2DEC271496446A326B22B7E08742   Satoshi/5=first
02 E8C7430A125F8FDA5B3AEA1C2C6CF7E404EE902B7150FE18CE397F2B4759AEC8   (Satoshi/5)*2
03 E1A4232B81AEA764680A74DC0F6CF58B26CB61048F0F70923DFD2F199B3C986C   ((Satoshi/5)*2)*2=second
03 678AFDB0FE5548271967F1A67130B7105CD6A828E03909A67962E0EA1F61DEB6   first+second

See? I splitted the public key from the Genesis Block into the sum of two public keys, where the second one is four times bigger than the first one. That means, I reached 1/5th of that key, because (1/5)+(4/5)=1. And you can easily validate that on your side, you need only point doubling, and point addition, nothing else.

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.

Well, let's see that on some private keys:

Code:

SHA-256("garlonicon")=272fc6644fedff1a897d6034bed23f61859e99440ee699033307976590316723
272fc6644fedff1a897d6034bed23f61859e99440ee699033307976590316723   d
a16ff47a7662cc9ee84c4670f2f6d97924556ffe6c267ebe16e623cf335da22e   d/5=first
42dfe8f4ecc5993dd0988ce1e5edb2f38dfc031629045d406df9e9119685031b   (d/5)*2
85bfd1e9d98b327ba13119c3cbdb65e71bf8062c5208ba80dbf3d2232d0a0636   ((d/5)*2)*2=second
272fc6644fedff1a897d6034bed23f61859e99440ee699033307976590316723   first+second

See? It works. Of course, obviously, if some private key is not divisible by 5, then you won't reach smaller key, you could get some bigger one as well. However, all ECDSA equations will be preserved in a given modulo.

[digaran]

obviously, if some private key is not divisible by 5, then you won't reach smaller key, you could get some bigger one as well. However, all ECDSA equations will be preserved in a given modulo.

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 :

Code:

02228c3cbe2d11af48c3b98bc2955cd2c293375f748e9aea9040ffba42c3f6be1e

When I subtract it from this key :

Code:

0000000000000000000000000000000000000400000000000000000000000000

I get this result :  *first result

Code:

0352a1087d5f64b39c33d54e02f678c50955770b736e2c7c646d0ee04e28fc83e7

Now here is another key :

Code:

038840d5318a6369971b6610cbef8a7cb998d8e106534881da28aabfefeec2f41b

If we subtract the above from this one :

Code:

0000000000000000000000000000000000000080000000000000000000000000

We will reach this key :  which is 1/8 of *first result.

Code:

037f32d1fbba6ec4ed91109538ea93afc9bb4f13e3052d319263c7bb05fcb109a7

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

[garlonicon]

Here is a problem I'm facing, it seems really easy to figure out, but I can't.

Because there is no solution. You can always divide by 5. In 20% cases, you will reach smaller key, in 80% some bigger one, just because every fifth number is divisible by 5. You can always divide by any number in range from 1 to n-1. Always.

You can clearly see that when trying to reach 1/2. In real numbers, it is some value between zero and one. But in ECDSA world, it could be for example 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1. You won't get any error, like "this division is impossible". It will never be rounded down to zero, or rounded up to one.

Division, exactly in the same way as multiplication, is always possible for all numbers from 1 to n-1. If you explore it on some smaller values, for example n=67 from my avatar, then you can clearly see that. You will rotate like in a clock, always getting "some" result, and never getting to the point of "no answer". And this is the beauty of the finite field, this is what makes it strong, and when you understand it, you will see, why you cannot "just divide it".

[GR Sasa]

Theoretically, if we could find a big number that by dividing it to the public key resulting in an exact number, we could solve the private key from public key.

But that's way too hard i guess, it's like guessing a whole 254-255 bits..

[digaran]

Here is a problem I'm facing, it seems really easy to figure out, but I can't.

Because there is no solution. You can always divide by 5. In 20% cases, you will reach smaller key, in 80% some bigger one, just because every fifth number is divisible by 5. You can always divide by any number in range from 1 to n-1. Always.

You can clearly see that when trying to reach 1/2. In real numbers, it is some value between zero and one. But in ECDSA world, it could be for example 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1. You won't get any error, like "this division is impossible". It will never be rounded down to zero, or rounded up to one.

Division, exactly in the same way as multiplication, is always possible for all numbers from 1 to n-1. If you explore it on some smaller values, for example n=67 from my avatar, then you can clearly see that. You will rotate like in a clock, always getting "some" result, and never getting to the point of "no answer". And this is the beauty of the finite field, this is what makes it strong, and when you understand it, you will see, why you cannot "just divide it".

Here is what I found, I'm not a mathematician but this should work and it can bend elliptic curve or rather straighten it.

56900346571233247÷5 = 11380069314246649.4 + n/5 + n/4 = 11380069314246649, I just couldn't find any meaningful result before, but I have tried the above method and it works.

It seems we can just divide it after all, but this won't help much if private key is unknown, these things are like fire for a child when they see it for the first time.

I wanted to say a monkey when you perform a magic trick in front of him, have you seen how excited they get? Lol.

[COBRAS]

56900346571233247÷5 = 11380069314246649.4 + n/5 + n/4 = 11380069314246649, I just couldn't find any meaningful result before, but I have tried the above method and it works.

O, f,it is really work ??

[_Counselor]

Here is what I found, I'm not a mathematician but this should work and it can bend elliptic curve or rather straighten it.

56900346571233247÷5 = 11380069314246649.4 + n/5 + n/4 = 11380069314246649, I just couldn't find any meaningful result before, but I have tried the above method and it works.

It seems we can just divide it after all, but this won't help much if private key is unknown, these things are like fire for a child when they see it for the first time.

I wanted to say a monkey when you perform a magic trick in front of him, have you seen how excited they get? Lol.

n works like zero in real numbers, n/x = n, n+x = x and so on.

[GR Sasa]

Sofar, i figured out that you can divide any pub keys with infinite numbers, and still get infinite valid results which goes back to the original pub key which results in infinite amount of collisions which results in an a big blackhole loop, that will never let you escape from it.

That's why I love Math. I saw some guy describing this as POINT_AT_INFINITY which i really agree with. Very well and nice describing