HSCTF 2020 Crypto - Chonky E

 

Chonky E

Allen and Jason rely on two different cryptosystems to keep their information secure.

Allen uses the RSA cryptosystem, given by the following public key:
e = 91043118409828550796773745518585981151180206101005135117565865602978722878478494447048783557571813980525643725323377488249838860897784683927029906188947001149632101513367258267329961684034661252866484981926055087386190015432964608927947646476193251820354738640453947833718397360834701566765504916472450194494897616371452996381159817427887623703639133290358520498419049175941584678802701606995099241245926884172985004839801270005583030514286561971825047719421487004569752638468907609110285739083279629747310953086535889932550905065172805818862336335628248528993024112446002398466115161473573451161053837400091893285717
n = 156749047558583013960513267351769479915110440411448078412590565797031533622509813352093119636835511977253033854388466854142753776146092587825440445182008237325262012698034419137157047927918635897378973846177552961727126115560551970797370239385129543828686170774323306933202481728884019420422360360849592983818405154473369790181636472137741865440233383956571081122982223602667853668754338360008279002325576495573847568301584365514417593244726435632222027817410359417329310347952169273512510934251453361933794586716533950489973436393834189505450956622286216819440777162804798432330933357058175885674184582816364542591313

Jason uses the Schmidt-Samoa cryptosystem. Although a public key has not been recovered, we know that Allen and Jason share the same primes (p,q).

A ciphertext was found on Jason’s computer that reads: 16267540901004879123859424672087486188548628828063789528428674467464407443871599865993337555869530486241139138650641838377419734897801380883629894166353225288006148210453677023750688175192317241440457768788267270422857060534261674538755743244831152470995124962736526978165448560149498403762447372653982922113772190234143253450918953235222315161964539311032659628670417496174123483045439359846360048774164337257829398345686635091862306204455687347443958931441225500856408331795261329035072585605404416473987280037959184981453888701567175803979981461050532113072292714696752692872526424122826696681194705563391161137426703690900733706866842363055967856443765215723398555522126909749236759332964873221973970368877565410624895160438695006432021529071866881905134494489266801004903504121740435965696128048690741210812963902631391765192187570107372453917327060678806282122942318369245760773848604249664378721970318257356486696764545

What are the contents of this message?

Since both crypto use same p and q doesnt reveal a method to recover p or q or the plaintext somehow, second BIG hint is the name of the challenge :- chonkyE

Big E could lead to two attacks, and Boneh Durfee attack is more powerful of the two.

Here is the evergreen sage script for the same. I modified it a bit (by fixing annoying python2 print statements) to run on sage 9 (python3).

d = 4801820624110300567381264152630360984400101198006662778338105999190025449039653722546363948393959163699344836724430590700225590643966670154013435626235133

We can recover p, q from n, e, d. The most convenient way according to me is to use RSA module from Crypto.PublicKey

from Crypto.PublicKey import RSA
import gmpy2

n = 156749047558583013960513267351769479915110440411448078412590565797031533622509813352093119636835511977253033854388466854142753776146092587825440445182008237325262012698034419137157047927918635897378973846177552961727126115560551970797370239385129543828686170774323306933202481728884019420422360360849592983818405154473369790181636472137741865440233383956571081122982223602667853668754338360008279002325576495573847568301584365514417593244726435632222027817410359417329310347952169273512510934251453361933794586716533950489973436393834189505450956622286216819440777162804798432330933357058175885674184582816364542591313
e = 91043118409828550796773745518585981151180206101005135117565865602978722878478494447048783557571813980525643725323377488249838860897784683927029906188947001149632101513367258267329961684034661252866484981926055087386190015432964608927947646476193251820354738640453947833718397360834701566765504916472450194494897616371452996381159817427887623703639133290358520498419049175941584678802701606995099241245926884172985004839801270005583030514286561971825047719421487004569752638468907609110285739083279629747310953086535889932550905065172805818862336335628248528993024112446002398466115161473573451161053837400091893285717
d = 4801820624110300567381264152630360984400101198006662778338105999190025449039653722546363948393959163699344836724430590700225590643966670154013435626235133

rsa = RSA.construct((n, e, d))
p = rsa.p
q = rsa.q

# rest is just implementation of Schmidt-Samoa

ct = 16267540901004879123859424672087486188548628828063789528428674467464407443871599865993337555869530486241139138650641838377419734897801380883629894166353225288006148210453677023750688175192317241440457768788267270422857060534261674538755743244831152470995124962736526978165448560149498403762447372653982922113772190234143253450918953235222315161964539311032659628670417496174123483045439359846360048774164337257829398345686635091862306204455687347443958931441225500856408331795261329035072585605404416473987280037959184981453888701567175803979981461050532113072292714696752692872526424122826696681194705563391161137426703690900733706866842363055967856443765215723398555522126909749236759332964873221973970368877565410624895160438695006432021529071866881905134494489266801004903504121740435965696128048690741210812963902631391765192187570107372453917327060678806282122942318369245760773848604249664378721970318257356486696764545

# N = p**2*q
# d = N^-1 mod lcm(p-1, q-1)
# m = c^d mod pq

N = p*q*q  # since this one works
lcm_val = gmpy2.lcm(p-1, q-1)
D = gmpy2.invert(N, lcm_val)
m = pow(ct, D, p*q)
print(bytes.fromhex(hex(m)[2:]))
# flag{remarkably_superb_acronym}