Rating:
Santa tried using public key cryptography to secure his letters, but his plan turned out to be a failure! After deleting his private keys, one of his elves found an encrypted letter from a child named Robin. Asking Robin what he wants for Christmas would be unprofessional, so the elf asked you to decrypt the letter. He also gave you access to a more basic copy of Santa's service.
Help Santa decrypt Robin's letter and you will definitely not be on the 'naughty' list this year!
Remote server: nc challs.xmas.htsp.ro 10005
Find n and e, realize e=2 means it's not RSA. Find Rabin's Cryptosystem and decrypt.
Connecting to the provided server shows the following prompt:
$ nc challs.xmas.htsp.ro 10005
__ ___ __
(_ /\ |\ | | /\ / (_
__) /--\ | \| | /--\ __)
_ ___ ___ _ _
| |_ | | |_ |_)
|_ |_ | | |_ | \
__ _ _ ___ _ _
(_ |_ |_) \ / | / |_
__) |_ | \ \/ _|_ \_ |_
Welcome to Santa's letter encryption service!
Enter your letter here and get an encrypted version which only Santa can read.
Note: Hackers beware: are using military-grade encryption!
Menu:
1. Encrypt a letter
2. Encrypt your favourite number
3. Exit
1337. Get Robin's Encrypted Letter (added by elf)
By playing around with encrypted numbers you quickly find small inputs are being squared:
Choice: 2
Your favourite number: 10
You can share your favourite number by sending him the following number: 100
But sufficiently large inputs are getting reduced modulo some number n:
Your favourite number: 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
You can share your favourite number by sending him the following number: 5245011011602491803664370575472709624274548272259320091433809466021203426817310186896310029658155776937528355323344006923977113553826210044716916761943666682385319625490349573808126763817318295478525328773318638213052114802345748146506404691689995405503303822016515046458252373127745904567619723597957297946422902129529543502645693599861443318116217613781848013941342646655217928664426579486750104198510693970393278733645664236923097146527594991807640347114720428789174163906930670960136912483150762248581418744755856956008035040747559729722485249549016117749161435076726496078746126920044963596268775441573974424006
There's several ways to find the value of n but I wrote a binary search:
def search():
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect(('challs.xmas.htsp.ro', 10005))
s.recv(1024)
high = gmpy2.mpz(10**309)
low = gmpy2.mpz(10**308)
while high > low:
print('Low: ', low)
print('High:', high)
print()
mid = (high + low) // 2
s.send(b'2\n')
s.recv(1024)
s.send(str(mid).encode('ascii') + b'\n')
response = s.recv(1024)
first_line = codecs.decode(response, 'ascii').split('\n')[0]
encrypted = int(first_line.split(' ')[-1])
if gmpy2.square(mid) == encrypted:
low = mid
else:
high = mid
Once it's converged, calculate n = high**2 - cipher(high). n happened to be
17150948086006853589591993610767711903029749167719666894975279147137565458158322238156960171902445590052801235253045792984069360111140927413022122124794074712372666903008787076313652986830735891457266804676322092444602549744787142273336096470832931113698218899620404912992099097015038863714351384075897287966440984446042594077540591489657561322101444523900312965276873402643875552954061610698504308548301539759131150366661281651087532807818489741520557925049746199503634928208501195328273501508911193754334803125090416259379171809642283452935819219835361791073290320084884025929676790915171638558685021113076310785793
which factors into two primes (found on FactorDB) as you'd expect with RSA, but e=2 has no inverse mod lambda(n) = (p-1)(q-1). For a while I read too much into the description and assumed it was incorrectly implemented RSA, but a hint from an organizer pointed me to look elsewhere. Searching something like "squared RSA cipher" in Google found Rabin's Cryptosystem, which lets you decrypt the message.
Flag: X-MAS{4cTu4lLy_15s_Sp3lLeD_r4b1n_n07_r0b1n_69316497123aaed43fc0}
