Matemith - CryptoCTF
For this challenge, we are given a matemith.sage and a output.txt. The challenge starts by splitting the flag into 6 parts (the number of parts can be later seen in the code):
l, flag = 14, flag.lstrip(b'CCTF{').rstrip(b'}') FLAG = [flag[l * i:l * (i + 1)] for i in range(len(flag) // l)] M = [bytes_to_long(_) for _ in FLAG] Then, 6 polynomials in $\mathbb{Q}[u, v, w, x, y, z]$ are generated:
p = getPrime(313) R.<u, v, w, x, y, z> = PolynomialRing(QQ) COEFS = [getRandomRange(1, p - 1) for _ in range(21)] f = COEFS[0] * u * v + COEFS[1] * u + COEFS[2] * v g = COEFS[3] * u * x * y + COEFS[3] * x + COEFS[5] * y + COEFS[6] * v h = COEFS[7] * u * w + COEFS[8] * w + COEFS[9] * u i = COEFS[10] * v * y * z + COEFS[11] * y + COEFS[12] * z + COEFS[13] * w j = COEFS[14] * v * w + COEFS[15] * v + COEFS[16] * w k = COEFS[17] * w * z * x + COEFS[18] * z + COEFS[19] * x + COEFS[20] * u f, g, h, i, j, k = R(f), R(g), R(h), R(i), R(j), R(k) Then, they are evaluated in the flag parts and printed:
