easyRSA - DASCTF 2022

Challenge

from Crypto.Util.number import *
p = 8960971163542317732496461517129968465242650917062063990499462344364441584827634827292160405689825408968674515002521127726347710721596547564600747499926623
q = 8760559466173555759649819558645388900176300543749720303611060081854268051122126311823383627928101599060554517177350782978533031317261512384261153195139913
e = 16875896916459
n = p * q
m = b'DASCTF{xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx}'
c = pow(bytes_to_long(m), e, n)
print(c)
#c = 46969729390923163243108804318186037502836930886558479407914362634132700908726538448687931567325086970674968889802077565679877050518631115039358482496465959812073755018022034189450875105664230732477034498644216866798324432699671109130715471019824874353986136599743110130002785245541745222681809172420644026934

Solve

Since gcd(e, p-1) = 1, we can just solve for m mod p.

\(c \equiv m^e\) (mod \(p\))

from Crypto.Util.number import *
p = 8960971163542317732496461517129968465242650917062063990499462344364441584827634827292160405689825408968674515002521127726347710721596547564600747499926623
q = 8760559466173555759649819558645388900176300543749720303611060081854268051122126311823383627928101599060554517177350782978533031317261512384261153195139913
e = 16875896916459
n = p * q
c = 46969729390923163243108804318186037502836930886558479407914362634132700908726538448687931567325086970674968889802077565679877050518631115039358482496465959812073755018022034189450875105664230732477034498644216866798324432699671109130715471019824874353986136599743110130002785245541745222681809172420644026934
assert gcd(e, p-1) == 1
m = mod(c, p).nth_root(e)
print(long_to_bytes(int(m)))
#DASCTF{1d86d07076bff93136ac60bbe61274d7}