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UIUCTF 2025 - Crypto Challenge Writeup
I’ve designed the shortest crypto challenge - can you find the flag?
author: epistemologist
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? Challenge Overview
We are provided with a Python source file (Download File):
from Crypto.Cipher import AES
from hashlib import md5
from secret import a,b,c,d, FLAG
assert a**4 + b**4 == c**4 + d**4 + 17 and max(a,b,c,d) < 2e4 and AES.new( f"{a*b*c*d}".zfill(16).encode() , AES.MODE_ECB).encrypt(FLAG).hex() == "41593455378fed8c3bd344827a193bde7ec2044a3f7a3ca6fb77448e9de55155"
Solution:
from Crypto.Cipher import AES
# Given ciphertext
ciphertext = bytes.fromhex("41593455378fed8c3bd344827a193bde7ec2044a3f7a3ca6fb77448e9de55155")
LIMIT = 3000 # Adjust based on available compute resources
# Step 1: Precompute fourth powers to avoid repeated exponentiation
fourth = [i**4 for i in range(LIMIT)]
# Step 2: Build all (c,d) pairs such that c^4 + d^4 = val
cd_sums = {}
for c in range(1, LIMIT):
for d in range(c, LIMIT): # symmetry: d ≥ c
val = fourth[c] + fourth[d]
cd_sums[val] = (c, d)
# Step 3: Try all (a, b) combinations
for a in range(1, LIMIT):
for b in range(a, LIMIT):
lhs = fourth[a] + fourth[b]
rhs_target = lhs - 17
if rhs_target in cd_sums:
c, d = cd_sums[rhs_target]
product = a * b * c * d
keystr = str(product).zfill(16)
if len(keystr) != 16:
continue # AES key must be exactly 16 bytes
key = keystr.encode()
cipher = AES.new(key, AES.MODE_ECB)
pt = cipher.decrypt(ciphertext)
try:
decoded = pt.decode()
if decoded.startswith("uiuctf{") and decoded.endswith("}"):
print(f"[+] ? Flag found: {decoded}")
print(f"[+] ? Key: {keystr}")
print(f"[+] ? Params: a={a}, b={b}, c={c}, d={d}")
exit()
except:
continue
