# Curve12833227 (crypto, 150+37p, 12 solved)
In the challenge we get [code](vuln.py) and the [result](flag.enc).
What the code does is encrypt the flag using AES and the AES key is encrypted via a custom elliptic curve using generator point `(4,10)`.We don't get the curve equation, only the addition and multiplication functions, but we can easily derive the curve parameters from that.
It's very simple to do from the point doubling part:
```pythonif u == w:     m = (3*u*w + 4*u + 1) * i(v+x)```
Of course since `u==w` then `3*u*w + 4*u + 1 = 3u^2 + 4u + 1`.Doubling a point on an elliptic curve requires calculating a tangent to the curve in this point, which means calculating a first derivative.
Since `3x^2 + 4x + 1` is the derivative we can easily integrate it to get `x^3 + 2x^2 + x +C` as the curve equation.
So we know that `y^2 = x^3 + 2x^2 + x + C`.
We also know that point `(4,10)` is on the curve, so we can apply this point to the curve to calculate `C = 100 - (64 + 2*16 + 4) = 0`.
So finally the curve is `y^2 = x^3 + 2x^2 + x = x(x^2+2x+1) = x(x+1)^2`
The issue now is that this curve has a singularity in `(-1,0)`.We follow the approach similar to: https://crypto.stackexchange.com/questions/61302/how-to-solve-this-ecdlp
And we get the solution code:
```pythonp = 2^128 - 33227P.<x> = GF(p)[]f = x^3 + 2*x^2 + xP = (4, 10)Q = (104708042197107879674895393611622483404, 276453155315387771858614408885950682409)
f_ = f.subs(x=x-1)print f_.factor() # 340282366920938463463374607431768178228
P_ = (P[0] +1, P[1])Q_ = (Q[0] +1, Q[1])
t = GF(p)(340282366920938463463374607431768178228).square_root()u = (P_[1] + t*P_[0])/(P_[1] - t*P_[0]) % pv = (Q_[1] + t*Q_[0])/(Q_[1] - t*Q_[0]) % p
print v.log(u)```
Which after a while gives us the discrete logarithm of `35996229751200732572713356533972460509`.
With this we can decrypt the flag:
```pythonk = 35996229751200732572713356533972460509aes = AES.new(long_to_bytes(k).ljust(16, '\0'), AES.MODE_CBC, '\0'*16)flag = "202bb05919b6f021d22a8baa7979ef6810761eb4c653b0fc5eebf2bc6ac6ecb052f887eedd075174abd884f84547df2d".decode("hex")print(len(flag))plaintext = aes.decrypt(flag)print(plaintext)```
And we get: `hxp{51n9uL4r1ti3s_r3duC3_tH3_G3nUs}`