LIT CTF 2024
simple otp
just xor with the given key, easy peasy
import random
from pwn import xor
encoded_with_xor = b'\x81Nx\x9b\xea)\xe4\x11\xc5 e\xbb\xcdR\xb7\x8f:\xf8\x8bJ\x15\x0e.n\\-/4\x91\xdcN\x8a'
random.seed(0)
key = random.randbytes(32)
print(xor(encoded_with_xor, key))
# LITCTF{sillyOTPlol!!!!sdfsgvkhf}
privatekey
N = 91222155440553152389498614260050699731763350575147080767270489977917091931170943138928885120658877746247611632809405330094823541534217244038578699660880006339704989092479659053257803665271330929925869501196563443668981397902668090043639708667461870466802555861441754587186218972034248949207279990970777750209
e = 89367874380527493290104721678355794851752244712307964470391711606074727267038562743027846335233189217972523295913276633530423913558009009304519822798850828058341163149186400703842247356763254163467344158854476953789177826969005741218604103441014310747381924897883873667049874536894418991242502458035490144319
c = 71713040895862900826227958162735654909383845445237320223905265447935484166586100020297922365470898490364132661022898730819952219842679884422062319998678974747389086806470313146322055888525887658138813737156642494577963249790227961555514310838370972597205191372072037773173143170516757649991406773514836843206
We can do the Boneh-Durfee Attack
import itertools
from Crypto.Util.number import long_to_bytes
def bivariate(f, bounds, m=1, d=None):
if d is None:
d = f.degree()
R = f.base_ring()
N = R.cardinality()
f_ = (f // f.lc()).change_ring(ZZ)
f = f.change_ring(ZZ)
l = f.lm()
M = []
for k in range(m+1):
M_k = set()
T = set((f^(m-k)).monomials())
for mon in (f^m).monomials():
if mon//l^k in T:
for extra in itertools.product(range(d), repeat=f.nvariables()):
g = mon * prod(map(power, f.variables(), extra))
M_k.add(g)
M.append(M_k)
M.append(set())
shifts = Sequence([], f.parent())
for k in range(m+1):
for mon in M[k] - M[k+1]:
g = mon//l^k * f_^k * N^(m-k)
shifts.append(g)
B, monomials = shifts.coefficients_monomials()
factors = [monomial(*bounds) for monomial in monomials]
for i, factor in enumerate(factors):
B.rescale_col(i, factor)
B = B.dense_matrix().LLL()
B = B.change_ring(QQ)
for i, factor in enumerate(factors):
B.rescale_col(i, 1/factor)
B = B.change_ring(ZZ)
H = [h for h in B * monomials if not h.is_zero()]
for f1, f2 in itertools.combinations(H, r=2):
x, y = f.parent().gens()
x = f1.parent()(x)
y = f1.parent()(y)
res = f1.resultant(f2,y).univariate_polynomial()
if res == 0:
continue
rs = res.roots()
if rs:
x = rs[0][0]
y = f1.subs(x=x).univariate_polynomial().roots()
if y != []:
return (x, y[0][0])
N = 91222155440553152389498614260050699731763350575147080767270489977917091931170943138928885120658877746247611632809405330094823541534217244038578699660880006339704989092479659053257803665271330929925869501196563443668981397902668090043639708667461870466802555861441754587186218972034248949207279990970777750209
e = 89367874380527493290104721678355794851752244712307964470391711606074727267038562743027846335233189217972523295913276633530423913558009009304519822798850828058341163149186400703842247356763254163467344158854476953789177826969005741218604103441014310747381924897883873667049874536894418991242502458035490144319
c = 71713040895862900826227958162735654909383845445237320223905265447935484166586100020297922365470898490364132661022898730819952219842679884422062319998678974747389086806470313146322055888525887658138813737156642494577963249790227961555514310838370972597205191372072037773173143170516757649991406773514836843206
A = (N+1)//2
PR.<x, y> = PolynomialRing(Zmod(e), 2)
f = x*(A+y) + 1
delta = .10 # guess
roots = bivariate(f, (2*floor(N^delta), isqrt(N)), m=2, d=2)
d = f.change_ring(ZZ)(*roots)/e
flag = pow(c, d, N)
print(long_to_bytes(flag))
# LITCTF{w13n3r_15_4n_unf0rtun4t3_n4m3}
pope shuffle
ct = [ord(i) for i in '࠴࠱࠼ࠫ࠼ࡣࡋࡍࠨ࡛ࡍ࡚ࡇ࡛ࠩࡔࡉࡌࡥ']
key = ord('L') - ct[0]
print(''.join([chr(i+key) for i in ct]))
# LITCTF{ce@ser_sAlad}
Symmetric RSA
#!/usr/bin/env python3
from Crypto.Util.number import long_to_bytes as ltb, bytes_to_long as btl, getPrime
p = getPrime(1024)
q = getPrime(1024)
n = p*q
e = p
with open("flag.txt", "rb") as f:
PT = btl(f.read())
CT = pow(PT, e, n)
print(f"{CT = }")
for _ in range(4):
CT = pow(int(input("Plaintext: ")), e, n)
print(f"{CT = }")
We will use this property which is true for all abs(x)<p:
>>> pow(x, p, n) % p == x
True
Then we can get p with gcd!
from Crypto.Util.number import long_to_bytes
from math import gcd
from pwn import remote
io = remote('litctf.org', 31783)
ct = int(io.recvline().decode().split()[-1])
out = []
for x in [2, 3, 5, 7]:
io.recv()
io.sendline(str(x).encode())
out.append(x - int(io.recvline().decode().split()[-1]))
p = gcd(*out)
flag = ct % p
print(long_to_bytes(flag))
# LITCTF{ju57_u53_e=65537_00a144ca}
Truly Symmetric RSA
#!/usr/bin/env python3
from Crypto.Util.number import long_to_bytes as ltb, bytes_to_long as btl, getPrime
p = getPrime(1536)
q = getPrime(1024)
n = p*q
e = p
with open("flag.txt", "rb") as f:
PT = f.read()
CT = pow(btl(PT), e, n)
print(f"{len(PT) = }")
print(f"{CT = }")
print(f"{n = }")
len(PT) = 62
CT = 155493050716775929746785618157278421579720146882532893558466000717535926046092909584621507923553076649095497514130410050189555400358836998046081044415327506184740691954567311107014762610207180244423796639730694535767800541494145360577247063247119137256320461545818441676395182342388510060086729252654537845527572702464327741896730162340787947095811174459024431128743731633252208758986678350296534304083983866503070491947276444303695911718996791195956784045648557648959632902090924578632023471001254664039074367122198667591056089131284405036814647516681592384332538556252346304161289579455924108267311841638064619876494634608529368113300787897715026001565834469335741541960401988282636487460784948272367823992564019029521793367540589624327395326260393508859657691047658164
n = 237028545680596368677333357016590396778603231329606312133319254098208733503417614163018471600330539852278535558781335757092454348478277895444998391420951836414083931929543660193620339231857954511774305801482082186060819705746991373929339870834618962559270938577414515824433025347138433034154976346514196324140384652533471142168980983566738172498838845701175448130178229109792689495258819665948424614638218965001369917045965392087331282821560168428430483072251150471592683310976699404275393436993044069660277993965385069016086918288886820961158988512818677400870731542293709336997391721506341477144186272759517750420810063402971894683733280622802221309851227693291273838240078935620506525062275632158136289150493496782922917552121218970809807935684534511493363951811373931
We have to use coppersmith for this one:
from Crypto.Util.number import long_to_bytes
CT = 155493050716775929746785618157278421579720146882532893558466000717535926046092909584621507923553076649095497514130410050189555400358836998046081044415327506184740691954567311107014762610207180244423796639730694535767800541494145360577247063247119137256320461545818441676395182342388510060086729252654537845527572702464327741896730162340787947095811174459024431128743731633252208758986678350296534304083983866503070491947276444303695911718996791195956784045648557648959632902090924578632023471001254664039074367122198667591056089131284405036814647516681592384332538556252346304161289579455924108267311841638064619876494634608529368113300787897715026001565834469335741541960401988282636487460784948272367823992564019029521793367540589624327395326260393508859657691047658164
n = 237028545680596368677333357016590396778603231329606312133319254098208733503417614163018471600330539852278535558781335757092454348478277895444998391420951836414083931929543660193620339231857954511774305801482082186060819705746991373929339870834618962559270938577414515824433025347138433034154976346514196324140384652533471142168980983566738172498838845701175448130178229109792689495258819665948424614638218965001369917045965392087331282821560168428430483072251150471592683310976699404275393436993044069660277993965385069016086918288886820961158988512818677400870731542293709336997391721506341477144186272759517750420810063402971894683733280622802221309851227693291273838240078935620506525062275632158136289150493496782922917552121218970809807935684534511493363951811373931
PR.<x> = PolynomialRing(Zmod(n))
flag = (x-CT).small_roots(X=256**62, beta=0.5)[0]
print(long_to_bytes(int(flag)))
# LITCTF{I_thought_the_bigger_the_prime_the_better_:(_72afea90}