### **[ORIGINAL WRITEUP](https://github.com/12flamingo/CTF-writeups/tree/main/Space%20Heroes%202022/%5Bcrypto%5D%20Easy%20Crypto%20Challenge): https://github.com/12flamingo/CTF-writeups/tree/main/Space%20Heroes%202022/%5Bcrypto%5D%20Easy%20Crypto%20Challenge**
---# [crypto]  Easy Crypto Challenge  (356 pt)Author: cdmann
Difficulty: Hard
## Challenge descriptionMy slingshotter cousin Maneo just discovered a new cryptography scheme and has been raving about it since. I was trying to tell him the importance of setting a **large, random private key**, but he wouldn't listen. Guess security isn't as important as how many thousands of credits he can win in his next race around the system.
I've recovered a message sent to him detailing the finish line. Can you decrypt the message to find the coordinates so I can beat him there? I'll give you a percentage of the score!
Enter flag as **shctf{x_y}**, where x & y are the coordinates of the decrypted point.
_Hint: You may wish to consult the great Sage of Crypto for help on this challenge._
### Files: * ecc.txt
---## Solution### TL;DR: Sage discrete_log to find private key### Recon* "large ... private key" - this means some sort of brute force might be involved :)* ecc.txt - could this stand for more than "**E**asy **C**rypto **C**hallenge"? Lacking some experience, I did not immediately recognise it, but a quick google search told me that it also stands for Elliptic Curve Cryptography! :smile: [[1]](https://en.wikipedia.org/wiki/Elliptic-curve_cryptography) > For crypto, an elliptic curve is a plane curve over a finite field **m**. This mean it is a set of integer coordinates within the field (a square matrix of m*m), satisfying the equation **y^2 = x^3 + ax + b (mod m)**  And indeed, we note the equation matches the one in our txt file. * "the great Sage of Crypto" - naturally, we turn to SageMath [[2]](https://sagecell.sagemath.org/)
**Quick overview of ECC**1. **Key Generation**: Maneo (Reciever) chooses a suitable curve `E(a,b)`, as well as a generator `G`. They also select an ideally "large, random private key" `d` for the encryption to be secure. The public key `P` is calculated with `P = d * G` --- (eq 1)2. **Encryption**: * Maneo sends `E(a, b), m, G, P` to sender. * Sender chooses a random number `k` in (0, m-1), and calculates point `(x1, y1) = k * G` --- (eq 2) and `(x2, y2) = k * P`.     * _Note: If (x2, y2) happen to be the [point at infinity](https://en.wikipedia.org/wiki/Point_at_infinity), sender chooses another k and recalculates._ * Ciphertext `C = (x3, y3) + (x2, y2)` --- (eq 3) where `(x3, y3)` is the **message** (that we are trying to steal for the challenge)* Finally, sender sends Maneo `C` and `(x1, y1)`3. **Decryption**: Maneo/Reciever calculates `(x2, y2) = k * P = k * d * G = d * (x1, y1) --- (eq 4).`  Then `(x3, y3) = C - (x2, y2)`
For a deeper dive and better understanding, I found this pretty informative: https://cryptobook.nakov.com/asymmetric-key-ciphers/elliptic-curve-cryptography-ecc [3]
### SolvingNow, given that we appear to be missing the private key `d`, we need to solve the Elliptic Curve Discrete Logarithm Problem (ECDLP) with `d` such that (eq 1) is satisfied. The challenge also tells us this key is unlikely to be "large" --> implying it must be small! (and thankfully, **Sage's discrete_log** function did work in our favour for this case). 
You can download SageMath, or use the online SageMathCell, as I have. 
_**sagemath**_``` python##INFO GIVEN & SET UPa = 3820149076078175358b = 1296618846080155687m = 11648516937377897327  #modulus F = FiniteField(m) #points in elliptic curve are integers within field E = EllipticCurve(F,[a,b]) #setting up curve function 
G = E(4612592634107804164, 6359529245154327104) #generator P = E(9140537108692473465, 10130615023776320406) #public key 
x1y1 = E(7657281011886994152, 10408646581210897023) # (x1, y1) = k * G --- (eq 2)C = E(5414448462522866853, 5822639685215517063) # C = (x3, y3) + (x2, y2) --- (eq 3)
##FINDING PRIVATE KEY d d = G.discrete_log(P) #such that P = d * G --- (eq 1)
##FINDING MESSAGE: (x3, y3)x2y2 = d * x1y1 # (x2, y2) = k * P = k * d * G = d * (x1, y1) --- (eq 4)x3y3 = C - x2y2 # --- (eq 3) print("x3:" + str(x3y3[0]))print("y3:" + str(x3y3[1]))```Output:```x3:8042846929834025144y3:11238981380437369357```
:triangular_flag_on_post: Hence, our flag is **shctf{8042846929834025144_11238981380437369357}**
## Additional notes/thoughtsWhile reading up on ECC, I also stumbled upon the Pohlig-Hellman attack [4] which>reduces discrete logarithm calculations to prime subgroups of the order of P and uses Chinese Remainder Theorem to solve system of congruences for discrete logarithm of the whole order. 
This would have been necessary if our numbers had been bigger (and factorisable into small primes)... 
### References
[1] https://en.wikipedia.org/wiki/Elliptic-curve_cryptography 
[2] https://sagecell.sagemath.org/ 
[3] https://cryptobook.nakov.com/asymmetric-key-ciphers/elliptic-curve-cryptography-ecc 
[4] https://hgarrereyn.gitbooks.io/th3g3ntl3man-ctf-writeups/content/2017/picoCTF_2017/problems/cryptography/ECC2/ECC2.html
