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Topic: Is there really 2^256 private key? (Read 75 times)

sr. member
Activity: 2618
Merit: 439
June 18, 2024, 12:28:55 AM
#5
2^256 is all the possible combinations of keys. But actually both ethereum and bitcoin uses this elliptical curve secp256k1. This elliptical curve denotes all the valid private keys.

The reason why 2^256 is not the maximum is because if it were the maximum, it would have already exceeded the range of the valid private keys in the elliptical curve. The range is n -1 wherein n is equal to 115792089237316195423570985008687907852837564279074904382605163141518161494336.

Last private key of eth is fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364140

Why not all ffffffff.......f (i,e 64 no of f or 2^256) ?

All the hexadecimal number (or those 0-9 and a-f characters) is the equivalent of n-1 so basically it is just ensuring that the range stays within the range of any point at the elliptical curve. If you surpass the range, it would be an invalid private key, which would result to an invalid public key and therefore cannot be used to obtain valid addresses.
sr. member
Activity: 448
Merit: 560
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June 16, 2024, 03:31:42 PM
#4
2²⁵⁶ is actually the number of possible outcome for a private key. Mathematically it's similar to also saying that the number of private keys that can exist is 2²⁵⁶ this is because every single outcome of a private key is possible to be an actual private key to a wallet.
The reason for such a huge number is because of the number of possible permutations for each and every space from a character, keeping in mind that the character could either be an alphabet or a number and since repetitions are possible the permutations become beven more. The permutation possible is the reason why it would be impossible to brute force such huge possiblities given that it would take forever for a regular computer to do so.
legendary
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June 16, 2024, 02:07:14 PM
#3
Why not all ffffffff.......f (i,e 64 no of f or 2^256) ?

Because of secp256k1.

Nearly every 256-bit number is a valid ECDSA private key. Specifically, any 256-bit number from 0x1 to 0xFFFF FFFF FFFF FFFF FFFF FFFF FFFF FFFE BAAE DCE6 AF48 A03B BFD2 5E8C D036 4140 is a valid private key.

The range of valid private keys is governed by the secp256k1 ECDSA standard used by Bitcoin.

The rest of the math is too much for me, but you can check this out if you are interested: https://en.bitcoin.it/wiki/Secp256k1


PS. I see now that you are asking about an altcoin, not Bitcoin. However, I'd guess that most (or all) of them use the same things/math as Bitcoin.
legendary
Activity: 3346
Merit: 3125
June 16, 2024, 01:52:59 PM
#2
There are total

115792089237316195423570985008687907852837564279074904382605163141518161494336

Private key in Etherium. It is less than 2^256

Last private key of eth is fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364140

Why not all ffffffff.......f (i,e 64 no of f or 2^256) ?

If you try to generate the ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff private key, you will get the error:

Error. Private Key is not in the 1,n-1 range

That means not all the numbers in the range of 0000000000000000000000000000000000000000000000000000000000000000 and ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff are valid for the private keys.

And a better question is:

Are there the same amount of address than private keys, the answer is no, there are more private keys than addresses which is something interesting but is another topic.
newbie
Activity: 20
Merit: 0
June 16, 2024, 12:28:30 PM
#1
There are total

115792089237316195423570985008687907852837564279074904382605163141518161494336

Private key in Etherium. It is less than 2^256

Last private key of eth is fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364140

Why not all ffffffff.......f (i,e 64 no of f or 2^256) ?
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