Blinding Light

RSA signing with malleability, we can sign anything but the admin token to retrieve the flag. My approach is to get the signature of the hex string b'\x02', in numeric value is $2^d$$2^d$, with $d$$d$ being the private key.

Then we request for the message that is twice the value of the admin token, then we get the value of $(2m{)}^d$$(2m)^d$. Then retrieving the value of $m^d$$m^d$ should be trivial.

xxxxxxxxxx
from Crypto.Util.number import bytes_to_long, long_to_bytes
from pwn import * 
import json 

admin_token = b'admin=True'

io = remote('socket.cryptohack.org', 13376)
io.recvline()

public_key = dict()
public_key['option'] = 'get_pubkey'

io.sendline(json.dumps(public_key).encode())

public_key = json.loads(io.recvline().decode())
N = int(public_key['N'], 16)
e = int(public_key['e'], 16)

two_times = b'\x02'.hex()
two_times_dict = dict()
two_times_dict['option'] = 'sign'
two_times_dict['msg'] = two_times
io.sendline(json.dumps(two_times_dict).encode())

two_times = json.loads(io.recvline().decode())
two_times = int(two_times['signature'], 16)

two_times_admin = long_to_bytes(2 * bytes_to_long(admin_token)).hex()
two_times_admin_dict = dict()
two_times_admin_dict['option'] = 'sign'
two_times_admin_dict['msg'] = two_times_admin

io.sendline(json.dumps(two_times_admin_dict).encode())
two_times_admin = json.loads(io.recvline().decode())
print(two_times_admin)
two_times_admin = int(two_times_admin['signature'], 16)

admin_sig = hex((two_times_admin * pow(two_times, -1, N)) % N)
admin = dict()
admin['option'] = 'verify'
admin['msg'] = admin_token.hex()
admin['signature'] = admin_sig

io.sendline(json.dumps(admin).encode())
io.interactive()

A smarter solution, by bobflanagan:

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# Preprocessing and preparation.
# 
# Express 'admin=True' as a long yields m=459922107199558918501733.
# Plugging this into a factorizer yields two prime factors:
#   p1=211578328037 and p2=2173767566209, m = p1*p2
#
# Asking the server to sign each of these messages individually will yield
# pow(p1, D, N) and pow(p2, D, N), and it will respond since neither of these
# individual messages is 'admin=True' when decoded.
#
# Now, you can compute the digital signature of 'admin=True' by multiplying the
# two signatures you got! Since p1^D * p2^D = (p1 * p2) ^ D mod N.

import json

from Crypto.Util.number import bytes_to_long, long_to_bytes
from pwn import *

p1 = 211578328037
p2 = 2173767566209

conn = remote('socket.cryptohack.org', 13376)
data = conn.recvline()

resp = dict()
resp['option'] = 'get_pubkey'
conn.send(json.dumps(resp))

data = json.loads(conn.recvline())
N = bytes_to_long(bytes.fromhex(data['N'][2:]))

resp = dict()
resp['option'] = 'sign'
resp['msg'] = long_to_bytes(p1).hex()
conn.send(json.dumps(resp))

data = json.loads(conn.recvline())
s1 = bytes_to_long(bytes.fromhex(data['signature'][2:]))

resp = dict()
resp['option'] = 'sign'
resp['msg'] = long_to_bytes(p2).hex()
conn.send(json.dumps(resp))

data = json.loads(conn.recvline())
s2 = bytes_to_long(bytes.fromhex(data['signature'][2:]))

signature = (s1 * s2) % N

resp = dict()
resp['option'] = 'verify'
resp['msg'] = b'admin=True'.hex()
resp['signature'] = long_to_bytes(signature).hex()
conn.send(json.dumps(resp))

data = json.loads(conn.recvline())
print(data['response'])