When I wanted to request a vanity address from https://vanitypool.appspot.com/ it asked me for my public key. I tried my main address, but it said it was too short. What should I put there?

1 Answer 1


ThePiachu explains it like this:

The idea is, that you can generate a vanity address without even knowing what private key it will actually belong to. One party grabs a random ECDSA keypair and saves the private key for themselves. They later can issue a request for a vanity address to be created, by supplying the public key form that keypair, their desired pattern and so forth. A person takes with looking for the vanity address is required to brute force random ECDSA private keys, get their corresponding public keys, add them to the provided public key, and proceed normally with SHA, RIPEMD and base58 until they receive the desired pattern. Then they give the private key they found to the first person, which adds the two private keys in order to get their vanity address. The best thing about it - the person looking for the solution won't know what the resulting private key is! This means that you can outsource your vanity key generation without needing to trust any third party.

An example (available from gobittest website): We have a private key: 18E14A7B6A307F426A94F8114701E7C8E774E7F9A47E2C2035DB29A206321725 which maps to public key: 0450863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A 299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6

and say we want to find a pattern "166". One of the solutions takes a form of a private key B18427B169E86DE681A1A62588E1D02AE4A7E83C1B413849989A76282A7B562F mapping to public key: 049C95E0949E397FACCECF0FE8EAD247E6FD082717E4A4A876049FB34A9ADED110DFEA2EF691CC4 A1410498F4C312F3A94318CD5B6F0E8E92051064876751C8404

If we add the two public keys (like the person looking for the solution would do), we get a public key: 0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487 which is equivalent to this address: 166ev9JXn2rFqiPSQAwM7qJYpNL1JrNf3h

If we add the two private keys (like the person requesting the address would), we get: CA65722CD418ED28EC369E36CFE3B7F3CC1CD035BFBF6469CE759FCA30AD6D54 which maps to the same public key as the sum of the public keys, and thus - to the same address.

Since this is a model that requires basically no trust from any party, I've decided to create this "Vanity Pool" to enable people to easily outsource their vanity address creation, as well as enable people wanting to earn some Bitcoins to use their machines for something different from traditional mining.

The simplest option to get the keypair to get started with the vanitypool is to go to https://www.bitaddress.org/ and generate a new address. Note down the private key. Now switch the tab to "Wallet Details" and enter the private key. One of the resulting fields is the "Public Key (130 characters [0-9A-F])" which has exactly the required format that is needed for the vanity pool.

Once a solution is found just combine the result that is mailed to you with the private key you noted before and you can import it into your Bitcoin client.

  • 1
    How do I combine the private keys? Can I use this website? gobittest.appspot.com/VanitySum
    – lurf jurv
    Jan 4, 2013 at 16:24
  • Yes that will your base private key and the found solution. I'm still looking for an offline tool to do the computation as submitting the base private key to a website defeats the purpose of splitting the base key from the solution.
    – cdecker
    Jan 4, 2013 at 16:30
  • Found one, there's a keyconv tool included with vanitygen. It can generate a keypair and combine private keys.
    – lurf jurv
    Apr 24, 2013 at 13:24
  • 1
    @cdecker, you wrote "If we add the two public keys...". The public key is a coordinate pair on the elliptic curve, so I wanted to verfy my understanding here. As you used it here, "add" means to combine the two points according to the group function of the elliptic curve, which is to identify the third point on the curve that intersects the line formed by the two coordinate pairs the public keys represent, and then negate its Y-coordinate, right? Dec 7, 2015 at 1:32
  • Not exactly sure about the math here, but that is my understanding as well, yes.
    – cdecker
    Dec 7, 2015 at 11:51

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