Rating:
Data from pcap:
Then problem can be simplified as below.
```python=
import numpy as np
from numpy._typing import NDArray
from gmpy2 import mpz
from typing import Any
def mpow(a: NDArray[Any], e: int, p: mpz):
n = a.shape[0]
c: NDArray[Any] = np.identity(n, dtype=object) // mpz(1)
for i in range(e.bit_length(), -1, -1):
c = (c @ c) % p
if e & (1 << i):
c = (c @ a) % p
return c
p = 15021738631187083129
G = [[8250369131483783446, 7636514724226618982, 13561904710751379340,
8146037206883500564],
[4902910229653653467, 12309994425856150343, 6519431793637084561,
826765390730273580],
[2293357986500986599, 3690712216774893256, 11221411595572149696,
6189443780743823928],
[4526003812557694577, 7070010165244546540, 2773849016628162168,
2528664545234724542]]
gorder = 48565318389923
A = [[7124857170442510436, 7098435244666432569, 5016319597248544408,
3368576660349080409],
[12766102894582816521, 1075168896515869195, 14940230223165707555,
13504245377824841154],
[11005568719346427939, 6351117183900723198, 10504073690105660358,
765373031704381609],
[4023860825601216122, 3960044495923291791, 14945314474664374706,
7318239600236598642]]
B = [[1282048969185546964, 12109372559076385336, 11131078763121815969,
8203066951109241796],
[13898425777031025196, 3180683801097577414, 4159815335494956274,
12490188080505436469],
[14804946988255087504, 14855384714399608402, 3036218738870877096,
2244875398266900486],
[11575966103907984574, 12991978517385440219, 8781891936464379787,
3191347143541874873]]
iv = bytes.fromhex('b90605c70dc5e0d4d2ab458c93ca3e2c')
cipher = bytes.fromhex('d772586e1e1ff88b5fbdc2a2f742d6c8fc517acc1d76ce38052db1aa266850cd264232b7d90d283a1d237a6f97d6558f')
```
We have:
$$ A = G^a \ \text{or} \ B = G^b$$
Where $G$ is a generator matrix and $A$ , $B$ are the public matrices.
To decrypt the cipher text we need to find $a$. Its a discrete logarithm problem in $GL(n,p)$ which is a group of invertible $n\times n$ matrix over $GF(p)$.
The order is 46 bits , so dlog easily found using baby step gaint step algorithm.
a = 14694400667272
Now decrypt:
```python=
Key = mpow(B, a, p)
```
Decrypt the AES with this shared key to get the flag.
