<h1> King Exchange Challenge Writeup [Crypto] </h1>
Luckily, I got First blood for this challenge.
Note :: This may not be the intended solution, I solved this challenge mostly by assuming something and verifying them.
In this challenge, we were given with two files `server.py` and `output.txt`.
The Execution flow of `server.py` is quite simple, it generates a key using `Diffie-Hellman Key Exchange` and `encryptsthe flag using AES` with shared key.
The Group over which the DH key exchange done is not given exactly, but I assumed from addition functionand Multiplication function that Group is related to Elliptic Curves.
This is the code for Addition and Multiplication of points.
```python# p is a primedef add_points(P, Q):    return ((P[0]*Q[0]-P[1]*Q[1]) % p, (P[0]*Q[1]+P[1]*Q[0]) % p)
def multiply(P, n):    Q = (1, 0)    while n > 0:        if n % 2 == 1:            Q = add_points(Q, P)        P = add_points(P, P)        n = n//2    return Q```
Multiplication is done using Double and Add method.
Searching through various forms of the elliptic curves for a similar addition law, I found that [Edwards Curve](https://en.wikipedia.org/wiki/Edwards_curve) addition law is quite similar to the addition law used by add_points function.
Edwards Elliptic Curves are of the form `x^2 + y^2 = 1 + d*x^2*y^2. (d != 0 & 1)`
Edwards Curve Additon law --> `(x1, y1) + (x2, y2) = (x1y2 + x2y1) / (1 + dx1x2y1y2) ,  (y1y2 - x1x2) / (1 - dx1x2y1y2).`
if d = 0     Addition law becomes  --> `(x1, y1) + (x2, y2) = x1y2 + x2y1 ,  y1y2 - x1x2 ` Assuming points are represented as `(x, y)` in `server.py` and `P = (x1, y1), Q = (x2, y2)``add_points(P, Q)` function calculates new point as `(x3, y3) = x1x2 - y1y2 , x1y2 + x2y1.`
clearly, both the formulas doesn't match and also neutral element of Edwards Curve is `(0, 1)` whereas neutral element used in the multiply function is `(1, 0)`. 
The formulas do match  if we consider that points in `server.py` are represented as `(y, x)`. Assuming that points are represented in `(y, x)` form and `P = (y1, x1) , Q = (y2, x2).` `add_points(P, Q)` function calculates new point as `(y3, x3) = (y1y2 - x1x2, x1y2 + x2y1).`
Now, both formulas and neutral elements match exactly.
Even though both formulas match, I am not sure whether this approach is correct or not because representing points in this `((y,x))` wayis quite unusual and Another issue is that when we substitute d = 0 in Edwards Curve equation, it becomes, `x^2 + y^2 = 1` which is `not an Elliptic Curve` but an `Equation of the Circle`.
The definition of the Edwards Curve also specifies that d != 0.
Coming back to solution, As both formulas match assuming that points statisfy the equation `x^2 + y^2 = 1`. We can easily calculate the prime p using this information.As we were given three points `(generator g, A's PublicKey A, B's PublicKey B)` which statisfy the equation modulo `p`.
==> `gx^2 + gy^2 = 1 + k1*p, Ax^2 + Ay^2 = 1 + k2*p, Bx^2 + By^2 = 1 + k3*p.`
so, `GCD(gx^2 + gy^2 - 1, Ax^2 + Ay^2 - 1, Bx^2 + By^2 - 1)` will give us the required `prime p`, somtimes along with small factors.
After obtaining prime p, I had no idea on how to proceed further.
I realized after a failed attempt that the function `add_points(P, Q)` does `Complex number multiplication` if we consider `P = P[0] + P[1]*i, Q = Q[0] + Q[1]*i` and `multiply(P, n)` function does `Modular Complex Exponentiation`.
I had searched how will be the structure of Group with `Modular Complex Multiplication` as operation and found [this paper](https://www.researchgate.net/publication/319731501_COMPLEX_PUBLIC_KEY_CRYPTOSYSTEMS).After reading few pages and realizing that elements in `Complex Finite Field`(term used in the linked paper) statisfy most properties of Finite fields.I wanted to find the number of elements are in a Complex Finite field with prime p. Using my small brain I calculated itas `(p**2 - 1)`. The reasoning for this is,
`complex numbers are of form (x + y*i) and x & y belongs to {0, 1,...,p-1} which gives p**2 combinations and taking out (0, 0)from the combinations results in p**2 - 1.`
I have verified this by exponentiating random complex numbers to `(p**2 - 1)` and checking that result is identity element `(1, 0)`.
for the `p` value which our challenge uses, `(p**2 - 1)` i.e order has many small factors. `(factors < 2**20)`
We can use the `pohlig-hellman` attack over the `Complex finite field` to solve the `discrete log` first in sub groups with the help of bruteforce and Using`Chinese Remainder Theorem` to calculate the complete secret.
`solve.py` contains my naive implementation of pohlig hellman attack and other solution code.
My Failed (May be foolish) Attempt is I tried to convert Edwards curve with d = 0 to Weierstrass form, resulting Elliptic Curve isis a singular curve (node)  but [this attack](https://crypto.stackexchange.com/questions/61302/how-to-solve-this-ecdlp/61434) cannot work directly cause `a` as in (x^2*(x + a)) is not a square in GF(p). We have to use other transformation to transfer points to Multiplicative group. (Attack details were given in Section 2.10 of the Elliptic Curves: Number Theory and Cryptography 2nd Edition book by Washington).
