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# Windows XP/Server 2003 Keys
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By Endermanch and WitherOrNot
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## *The Problem*
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**In general, the only thing that separates us from generating valid Windows XP keys for EVERY EDITION and EVERY BUILD is the lack of respective private keys generated from their public counterparts inside `pidgen.dll`**. There's no code for the elliptic curve discrete logarithm function widely available online, there's only vague information on how to do it.
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As time went on, the problem has been _partially_ solved.
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The BINK resource was not encoded in any way and the data was just sequentially written to the resource. **sk00ter** also fully explained the BINK format on the MDL forums.
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Utilizing prior community knowledge on the subject, I wrote a BINK Reader in Python 3. The file is public in this repository, [click here](https://github.com/Endermanch/XPKeygen/blob/main/BINKReader.py) to view the source code.
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The discrete logarithm solution is the most unexplored area of research as of **May 28th, 2023**. However, my friend **nephacks** did find that elusive tool to solve that difficult problem in the darkest corners of the internet.
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It's called ECDLP (Elliptic Curve Discrete Logarithm Problem) Solver by Mr. HAANDI. Since it was extremely frustrating to find online, I did reupload it on my website. You can download the tool [here](https://dl.malwarewatch.org/software/advanced/ecc-research-tools/).
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The ReadMe file that comes with the version **0.2a** of the solver is good enough by itself, so anyone with a brain will be able to set that tool up. However, it's not open-source, so integrating it into my keygen is proven impossible.
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<details open>
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In the ideal scenario, the keygen would ask you for a BINK-resource extracted from `pidgen.dll`, which it would then unpack into the following segments:
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* Public key (`pubX`; `pubY`)
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* Generator (`genX`; `genY`)
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* Base point (`a`; `b`)
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* Point count `p`
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Knowing these segments, the keygen would bruteforce the geneator order `genOrder` using Schoof's algorithm followed by the private key `privateKey`, leveraging the calculated `genOrder` to use the most optimal Pollard's Rho algorithm. There's no doubt we can crack any private key in a matter of 20 minutes using modern computational power, provided we have the working algorithm.
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Once the keygen finishes bruteforcing the correct private key, the task boils down to actually generating a key, **which this keygen does**.
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To give you a better perspective, I can provide you with the flow of the ideal keygen. Crossed out is what my keygen implements:
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* ~~BINK resource extraction~~
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* Bruteforce Elliptic Curve discrete logarithm solution (`genOrder`, `privateKey`)
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* ~~Product Key processing mechanism~~
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* ~~Windows XP key generation~~
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* ~~Windows XP key validation~~
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* ~~Windows Server 2003 key generation~~
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</details>
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# Principle of operation
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We need to use a random Raw Product Key as a base to generate a Product ID in a form of `AAAAA-BBB-CCCCCCS-DDEEE`.
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## Product ID
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| Digits | Meaning |
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|-------:|:----------------------|
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| AAAAA | OS Family constant |
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| BBB | Channel ID |
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| CCCCCC | Sequence Number |
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| S | Check digit |
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| DD | Public key index |
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| EEE | Random 3-digit number |
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The OS Family constant `AAAAA` is different for each series of Windows XP. For example, it is 76487 for SP3.
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The `BBB` and `CCCCCC` sections essentially encode the Raw Product Key. For example, if the first section is equal to `XXX` and the second section is equal to `YYYYYY`, the Raw Product Key will be encoded as `XXX-YYYYYY`.
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The check digit `S` is picked so that the sum of all `C` digits with it added makes a number divisible by 7.
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The public key index `DD` lets us know which public key was used to successfully verify the authenticity of our Product Key.
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For example, it's `22` for Professional keys and `23` for VLK keys.
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A random number `EEE` is used to generate a different Installation ID each time.
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## Product Key
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The Product Key itself (not to confuse with the RPK) is in form `FFFFF-GGGGG-HHHHH-JJJJJ-KKKKK`, encoded in Base-24 with
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the alphabet `BCDFGHJKMPQRTVWXY2346789` to exclude any characters that can be easily confused, like `I` and `1` or `O` and `0`.
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As per the alphabet capacity formula, the key can at most contain 114 bits of information.
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$$N = \log_2(24^{25}) \approx 114$$
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Based on that calculation, we unpack the 114-bit Product Key into 4 ordered segments:
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| Segment | Capacity | Data |
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|-----------|----------|-------------------------------------------|
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| Upgrade | 1 bit | Upgrade version flag |
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| Serial | 30 bits | Raw Product Key (RPK) |
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| Hash | 28 bits | RPK hash |
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| Signature | 55 bits | Elliptic Curve signature for the RPK hash |
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For simplicity' sake, we'll combine `Upgrade` and `Serial` segments into a single segment called `Data`. By that logic we'll be able to extract the RPK by
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shifting `Data` right and pack it back by shifting bits left, because most a priori valid product keys I've checked had the Upgrade bit set to 1.
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Microsoft redid their Product Key format with Windows Server 2003 to include a backend server authentication key, which was an actually secure approach to
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license validation, as no one could ever make a guess on which validation algorithm they had employed on their private server. Besides adding the online
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validation mechanism, they also cranked up the overall arithmetic from 384 to 512 bits, and the signature scalar to 62 bits of information.
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| Segment | Capacity | Data |
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|------------|----------|-------------------------------------------|
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| Upgrade | 1 bit | Upgrade version flag |
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| Channel ID | 10 bits | The `BBB` part of the RPK |
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| Hash | 31 bits | RPK hash |
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| Signature | 62 bits | Elliptic Curve signature for the RPK hash |
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| Auth Key | 10 bits | Backend authentication value |
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However, if we generated a key without the online activation in mind, we still could generate valid keys that would let us through the setup of the operating system.
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And that's exactly what the code does - it generates a random 10-bit authentication key. Nowadays it doesn't matter at all, as activation servers are down and
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Server 2003 is considered abandonware, the same way this entire project shouldn't be considered piracy.
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## Elliptic Curves
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Elliptic Curve Cryptography (ECC) is a type of public-key cryptographic system.
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This class of systems relies on challenging "one-way" math problems - easy to compute one way and intractable to solve the "other" way.
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Sometimes these are called "trapdoor" functions - easy to fall into, complicated to escape.<sup>[5]</sup>
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ECC relies on solving equations of the form
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$$y^2 = x^3 + ax + b$$
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In general, there are 2 special cases for the Elliptic Curve leveraged in cryptography - **F<sub>2m</sub>** and **F<sub>p</sub>**.
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They differ only slightly. Both curves are defined over the finite field, F<sub>p</sub> uses a prime parameter that's larger than 3,
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F<sub>2m</sub> assumes $p = 2m$. Microsoft used the latter in their algorithm.
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An elliptic curve over the finite field F<sub>p</sub> consists of:
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* a set of integer coordinates ${x, y}$, such that $0 \le x, y < p$;
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* a set of points $y^2 = x^3 + ax + b \mod p$.
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**An elliptic curve over F<sub>17</sub> would look like this:**
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The curve consists of the blue points in above image. In practice the "elliptic curves"
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used in cryptography are "sets of points in a square matrix".
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The above curve is "educational". It provides very small key length (4-5 bits).
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In real world situations developers typically use curves of 256-bits or more.
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An important concept is that addition can be defined between two points on an elliptic curve. This also allows a definition of integer multiplication ($nP$ is $P$ added to itself $n$ times).
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The core of elliptic curve cryptography uses this multiplication definition, as solving the equation $Q=nP$ for known points $P$ and $Q$ is difficult with known algorithms, providing cryptographic security.
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## BINK resource
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Since it is a public-key cryptographic system, Microsoft had to share the public key with their Windows XP release to check entered product keys against.
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It is stored within `pidgen.dll` in a form of a BINK resource. The first set of BINK data is there to validate retail keys, the second is for the
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OEM keys respectively.
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**The structure of the BINK resource for Windows 98 and Windows XP is as follows:**
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| Offset | Value |
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|---------:|:---------------------------------------------------------------------|
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| `0x0000` | BINK ID |
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| `0x0004` | Size of BINKEY structure in bytes (always `0x16C` in practice) |
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| `0x0008` | Header length (always `7` in practice) |
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| `0x000C` | Checksum |
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| `0x0010` | Number-encoded date - BINKEY version (always `19980206` in practice) |
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| `0x0014` | ECC curve order size (always `12` in practice) |
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| `0x0018` | Hash length (always `28` in practice) |
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| `0x001C` | Signature length (always `55` in practice) |
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| `0x0020` | Finite Field Order `p` |
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| `0x005C` | Curve Parameter `a` |
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| `0x0098` | Curve Parameter `b` |
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| `0x00D4` | Base Point x-coordinate `Gx` |
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| `0x0110` | Base Point y-coordinate `Gy` |
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| `0x014C` | Public Key x-coordinate `Kx` |
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| `0x0188` | Public Key y-coordinate `Ky` |
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Each segment is marked with a different color, the BINK header values are the same.
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**Windows Server 2003 and Windows XP x64 implement it differently:**
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| Offset | Value |
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|---------:|:---------------------------------------------------------------------|
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| `0x0000` | BINK ID |
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| `0x0004` | Size of BINKEY structure in bytes |
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| `0x0008` | Header length (always `9` in practice) |
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| `0x000C` | Checksum |
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| `0x0010` | Number-encoded date - BINKEY version (always `20020420` in practice) |
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| `0x0014` | ECC curve order size (always `16` in practice) |
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| `0x0018` | Hash length (always `31` in practice) |
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| `0x001C` | Signature length (always `62` in practice) |
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| `0x0020` | Backend authentication value length (always `12` in practice) |
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| `0x0024` | Product ID length (always `20` in practice) |
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| `0x0028` | Finite Field Order `p` |
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| `0x0068` | Curve Parameter `a` |
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| `0x00A8` | Curve Parameter `b` |
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| `0x00E8` | Base Point x-coordinate `Gx` |
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| `0x0128` | Base Point y-coordinate `Gy` |
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| `0x0168` | Public Key x-coordinate `Kx` |
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| `0x01A8` | Public Key y-coordinate `Ky` |
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**And here are my structure prototypes made for the BINK Reader in C:**
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```c
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typedef struct _EC_BYTE_POINT {
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CHAR x[256]; // x-coordinate of the point on the elliptic curve.
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CHAR y[256]; // y-coordinate of the point on the elliptic curve.
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} EC_BYTE_POINT;
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typedef struct _BINKHDR {
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// BINK version - not stored in the resource.
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ULONG32 dwVersion;
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// Original BINK header.
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ULONG32 dwID;
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ULONG32 dwSize;
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ULONG32 dwHeaderLength;
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ULONG32 dwChecksum;
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ULONG32 dwDate;
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ULONG32 dwKeySizeInDWORDs;
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ULONG32 dwHashLength;
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ULONG32 dwSignatureLength;
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// Extended BINK header. (Windows Server 2003+)
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ULONG32 dwAuthCodeLength;
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ULONG32 dwProductIDLength;
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} BINKHDR;
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typedef struct _BINKDATA {
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CHAR p[256]; // Finite Field order p.
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CHAR a[256]; // Elliptic Curve parameter a.
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CHAR b[256]; // Elliptic Curve parameter b.
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EC_BYTE_POINT G; // Base point (Generator) G.
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EC_BYTE_POINT K; // Public key K.
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} BINKDATA;
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typedef struct _BINKEY {
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BINKHDR header;
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BINKDATA data;
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} BINKEY;
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```
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In case you want to explore further, the source code of `pidgen.dll` and all its functions is available within this repository, in the "pidgen" folder.
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## Validating / generating product keys
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Please note throughout this section that whenever integers are converted to bytes or vice-versa, they are in [little-endian](https://en.wikipedia.org/wiki/Endianness#Little) byte order.
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First, let's define some variables used in these algorithms:
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- $m$, the `Data` section of the product key (Raw Product Key/Channel ID and Upgrade bit)
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- $h$, the `Hash` section of the product key
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- $s$, the `Signature` section of the product key
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- $k$, the private key, used to generate product keys
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- $a$, the `Authentication Info` section of the product key (BINK2002 only)
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$G$, $K$, $p$, $a$, and $b$ are constants that come from the BINK, described above.
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The constant $n$, not included in the BINK, is the order of the point $G$.
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### BINK1998 (Windows 98/XP era)
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#### Validation
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1. Using base-24 conversion, with the base alphabet `BCDFGHJKMPQRTVWXY2346789`, convert the product key into an integer
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2. From this integer, extract the values $m$, $h$, and $s$ using bit-shifting and logical ANDs
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3. Compute the elliptic curve point $R = hK + sG$
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4. Compute `digest = SHA1(m || R.x || R.y)`, where `||` represents byte concatenation
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5. Convert `digest` to an integer
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6. Let $h_t$ be the upper 28 bits of `digest`
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7. Compare $h$ and $h_t$, if they are equal, the product key is valid
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#### Generation
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1. Compute a random number $c$
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2. Compute the random point $R = cG$
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3. Compute `digest = SHA1(m || R.x || R.y)`
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4. Let $h$ be integer formed by the lower 28 bits of `digest`
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5. Let $s = kh + c \pmod{p}$
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6. Pack $s$, $h$, and $m$ into a 114-bit integer
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7. Convert this integer into a product key, using base-24 conversion
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#### Mathematical mechanism
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The point $K$ is related to the point $G$ as $K=-kG$. Thus,
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$$ R = hK + sG = -hkG + \left(hk + c\right)G = cG $$
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The values $h$ and $s$ are dependent on the value of $R$ and $m$, so they can only be found by knowing the private key $k$.
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### BINK2002 (Windows Server 2003/Windows XP x64 era)
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#### Validation
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1. Using base-24 conversion, with the base alphabet `BCDFGHJKMPQRTVWXY2346789`, convert the product key into an integer
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2. From this integer, extract the values $m$, $h$, and $s$ using bit-shifting and logical ANDs
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3. Compute `digest1 = SHA1(5D || m || h || a || 00 00)`
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4. Let $e$ be the integer formed by the lower 62 bits of `digest1`
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5. Let $R = s(sG + eK)$
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6. Let `digest2 = SHA1(79 || m || R.x || R.y)`
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7. Let $h_t$ be the lower 31 bits of `digest2`
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8. Compare $h$ and $h_t$, if they are equal, the product key is valid
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Afterwards, the value $a$, along with some other information, can be sent to the activation server for further validation.
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#### Generation
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1. Compute a random number $c$
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2. Compute the random point $R = cG$
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3. Let `digest2 = SHA1(79 || m || R.x || R.y)`
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4. Let $h$ be integer formed by the lower 31 bits of `digest2`
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5. Let $s = \frac{-ek + \sqrt{\left(ek\right)^2 + 4c}}{2} \pmod {n}$
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6. Compute $a$ by unknown algorithm
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7. Pack $s$, $h$, and $m$, and $a$ into a 114-bit integer
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8. Convert this integer into a product key, using base-24 conversion
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#### Mathematical mechanism
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The variables used in this algorithm are related as follows:
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$$ R=s(sG + eK) = {s^2}G + seK = {s^2}G - sekG = cG $$
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$$ \left(s^2 - eks - c\right)G = O $$
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$$ s^2 - eks - c = 0 $$
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During generation, $s$ is found as the solution to the above quadratic equation.
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### Reversing the private key
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If we want to generate valid product keys for Windows XP, we must compute the corresponding private key using the public key supplied with `pidgen.dll`,
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which means we have to reverse-solve the one-way ECC task.
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Judging by the key located in BINK, the curve order is **384 bits** long in Windows XP and **512 bits** long in Server 2003 / XP x64 respectively.
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The computation difficulty using the most efficient Pollard's Rho algorithm with asymptotic complexity $O(\sqrt{n})$ would be at least $O(2^{168})$ for Windows XP, and $O(2^{256})$ for Windows Server 2003, but lucky for us,
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||||||
|
Microsoft limited the value of the signature to 55 bits in Windows XP and 62 bits in Windows Server 2003 in order to reduce the amount of matching product keys, reducing the difficulty to a far more manageable $O(2^{28})$ / $O(2^{31})$.
|
||||||
|
|
||||||
|
As mentioned before, there's only one public tool that satisfies our current needs, which is the ECDLP solver by Mr. HAANDI.<br>
|
||||||
|
|
||||||
|
To compute the private key, we will need to supply the tool with the public ECC values located in the BINK resource, as well as the order `genOrder` of the base point `G(Gx; Gy)`.
|
||||||
|
The order of the base point can be computed using SageMath.
|
||||||
|
|
||||||
|
**Here's the basic algorithm I used to reverse the Windows 98 private key:**
|
||||||
|
|
||||||
|
1. Compute the order of the base point using **SageMath**. In SageMath, execute the following commands:
|
||||||
|
1) `E = EllipticCurve(GF(p), [0, 0, 0, a, b])`, where `p`, `a` and `b` are decimally represented elliptic curve parameters from the BINK resource.
|
||||||
|
2) `G = E(Gx, Gy)`, where `Gx` and `Gy` are decimally represented base point coordinates from the BINK resource.
|
||||||
|
3) `K = E(Kx, Ky)`, where `Kx` and `Ky` are decimally represented public key coordinates from the BINK resource.
|
||||||
|
4) `n = G.order()`, `n` will be the computed order of the base point. **It may take some time to compute, even on the newest builds.**
|
||||||
|
5) Factor the order using `factor(n)`. Microsoft used prime numbers for the point orders, so if it returns the number itself, it's completely normal.
|
||||||
|
6) Save the resulting factors of the order somewhere.
|
||||||
|
7) `-K` will give you the inverse of the public key in a projective plane with coordinates `(x : y : z)`. Save the `y` coordinate somewhere, it is required to generate a correct private key.
|
||||||
|
2. Compute the private key using **ECDLP Solver v0.2a**.
|
||||||
|
1) The tool comes with a template job `job_template.txt` and a ReadMe file. It's necessary to understand how the tool works to use it.
|
||||||
|
2) Insert all public elliptic curve values from the BINK resource, **except the `Ky` coordinate**. To generate a correct private key, **you must use the inverse coordinate `-Ky` you have calculated in SageMath earlier.**
|
||||||
|
3) Insert the factors of the base point order `n` and specify the factor count. It will very likely be `1`, as Microsoft mainly uses primes for their generator orders.
|
||||||
|
4) Run the tool `<arch> ECDLP Solver.exe <job_name>.txt` and wait until it calculates the private key `k = %d` for you.
|
||||||
|
|
||||||
|
**Here's an example of the Windows XP job `job_xp.txt` that yields the correct private key for the ECDLP Solver.**
|
||||||
|
|
||||||
|
```pascal
|
||||||
|
GF := GF(22604814143135632990679956684344311209819952803216271952472204855524756275151440456421260165232069708317717961315241);
|
||||||
|
E := EllipticCurve([GF|1,0]);
|
||||||
|
G := E![10910744922206512781156913169071750153028386884676208947062808346072531411270489432930252839559606812441712224597826,19170993669917204517491618000619818679152109690172641868349612889930480365274675096509477191800826190959228181870174];
|
||||||
|
K := E![14399230353963643339712940015954061581064239835926823517419716769613937039346822269422480779920783799484349086780408,17120082747148185997450361756610881166187863099877353630300913555824935802439591336620545428308962346299700128114607];
|
||||||
|
/*
|
||||||
|
FactorCount:=1;
|
||||||
|
61760995553426173
|
||||||
|
*/
|
||||||
|
```
|
||||||
|
|
||||||
|
**And the ECDLP Solver output for it:**
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
## DPCDLL and Channel IDs
|
||||||
|
|
||||||
|
In Windows versions before XP, the channel ID is rarely validated except in certain products, such as Windows 98 SE Select Edition.
|
||||||
|
In Windows XP and Server 2003, however, the channel ID is validated to determine the license type for a specific copy.
|
||||||
|
This validation is done by DPCDLL.DLL, which contains a table of signed channel ID ranges and their associated license type.
|
||||||
|
|
||||||
|
For all NT 5 Windows versions except Server 2003 R2, this table has the following structure for each row:
|
||||||
|
|
||||||
|
| Offset | Value |
|
||||||
|
|----------|-----------------------------------|
|
||||||
|
| `0x0000` | Index |
|
||||||
|
| `0x0004` | BINK ID |
|
||||||
|
| `0x0008` | Minimum Channel ID |
|
||||||
|
| `0x000C` | Maximum Channel ID |
|
||||||
|
| `0x0010` | License Type |
|
||||||
|
| `0x0014` | Activation Grace Period (in days) |
|
||||||
|
| `0x0018` | Evaluation Period (in days) |
|
||||||
|
| `0x001C` | Signature Length |
|
||||||
|
| `0x0020` | Signature |
|
||||||
|
|
||||||
|
The license types are as follows:
|
||||||
|
|
||||||
|
| Type | Meaning |
|
||||||
|
|------|--------------------------------------------------|
|
||||||
|
| 1 | Volume (VLK) |
|
||||||
|
| 2 | Retail/OEM Certificate of Authenticity (OEM-COA) |
|
||||||
|
| 3 | Evaluation |
|
||||||
|
| 4 | Tablet |
|
||||||
|
| 5 | OEM System Locked Pre-Installation (OEM-SLP) |
|
||||||
|
| 6 | Embedded |
|
||||||
|
|
||||||
|
For the Evaluation Period and Activation Grace Period values, the value `2147483647` is used to indicate `N/A`.
|
||||||
|
|
||||||
|
When a key with an invalid channel ID and correct BINK is supplied during setup, the setup installed will accept the key since it has a valid signature.
|
||||||
|
However, when attempting to login to the system, the system will prevent login until the system is activated.
|
||||||
|
When attempting to activate, a non-functional OOBE Activation popup will appear, blocking login and effectively bricking the system, as shown below.
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
This strange behavior is most likely a bug triggered by a lack of error handling in DPCDLL's channel ID validation code.
|
||||||
|
|
||||||
|
## Literature
|
||||||
|
I will add more decent reads into the bibliography in later releases.
|
||||||
|
|
||||||
|
**Understanding basics of Windows XP Activation**:
|
||||||
|
* [[1] Inside Windows Product Activation - Fully Licensed](https://www.licenturion.com/xp/fully-licensed-wpa.txt)
|
||||||
|
* [[2] MSKey 4-in-1 ReadMe](https://malwarewatch.org/documents/MSKey4in1.pdf)
|
||||||
|
* [[3] Windows序列号产生原理(椭圆曲线法)](https://blog.csdn.net/zhiyuan411/article/details/5156330)
|
||||||
|
|
||||||
|
**Understanding Elliptic Curve Cryptography**:
|
||||||
|
* [[4] Elliptic Curve Cryptography for Beginners - Matt Rickard](https://matt-rickard.com/elliptic-curve-cryptography)
|
||||||
|
* [[5] Elliptic Curve Cryptography (ECC) - Practical Cryptography for Developers](https://cryptobook.nakov.com/asymmetric-key-ciphers/elliptic-curve-cryptography-ecc)
|
||||||
|
* [[6] A (Relatively Easy To Understand) Primer on Elliptic Curve Cryptography - Cloudflare](https://blog.cloudflare.com/a-relatively-easy-to-understand-primer-on-elliptic-curve-cryptography/)
|
||||||
|
|
||||||
|
**Public discussions**:
|
||||||
|
* [[7] Windows 98 Equivalent // Server 2003 Algorithm](https://github.com/Endermanch/XPKeygen/issues/3)
|
||||||
|
* [[8] Cracking Windows XP](https://forums.mydigitallife.net/threads/is-there-any-way-to-crack-decrypt-the-winxp-consumer-activation-system-to-generate-activation-ids.80133/)
|
||||||
Loading…
Reference in New Issue