cryptohack wp challenges (RSA篇) （持续更新）
一些没有的都在公钥密码学那篇找到
Inferius Prime
能分解n
p= 986369682585281993933185289261
q= 752708788837165590355094155871
e=3
c=39207274348578481322317340648475596807303160111338236677373
n=p*q
fn=(p-1)*(q-1)
print(n)
print(fn)
import libnum
import gmpy2
import sympy
d=gmpy2.invert(e,fn)
print("d=",d)
flag=int(pow(c,d,n))
print('flag=',flag)
print(libnum.n2s(flag))
print(sympy.nextprime(d))
Square Eyes
import gmpy2
import libnum
n = 535860808044009550029177135708168016201451343147313565371014459027743491739422885443084705720731409713775527993719682583669164873806842043288439828071789970694759080842162253955259590552283047728782812946845160334801782088068154453021936721710269050985805054692096738777321796153384024897615594493453068138341203673749514094546000253631902991617197847584519694152122765406982133526594928685232381934742152195861380221224370858128736975959176861651044370378539093990198336298572944512738570839396588590096813217791191895941380464803377602779240663133834952329316862399581950590588006371221334128215409197603236942597674756728212232134056562716399155080108881105952768189193728827484667349378091100068224404684701674782399200373192433062767622841264055426035349769018117299620554803902490432339600566432246795818167460916180647394169157647245603555692735630862148715428791242764799469896924753470539857080767170052783918273180304835318388177089674231640910337743789750979216202573226794240332797892868276309400253925932223895530714169648116569013581643192341931800785254715083294526325980247219218364118877864892068185905587410977152737936310734712276956663192182487672474651103240004173381041237906849437490609652395748868434296753449
e = 65537
ct = 222502885974182429500948389840563415291534726891354573907329512556439632810921927905220486727807436668035929302442754225952786602492250448020341217733646472982286222338860566076161977786095675944552232391481278782019346283900959677167026636830252067048759720251671811058647569724495547940966885025629807079171218371644528053562232396674283745310132242492367274184667845174514466834132589971388067076980563188513333661165819462428837210575342101036356974189393390097403614434491507672459254969638032776897417674577487775755539964915035731988499983726435005007850876000232292458554577437739427313453671492956668188219600633325930981748162455965093222648173134777571527681591366164711307355510889316052064146089646772869610726671696699221157985834325663661400034831442431209123478778078255846830522226390964119818784903330200488705212765569163495571851459355520398928214206285080883954881888668509262455490889283862560453598662919522224935145694435885396500780651530829377030371611921181207362217397805303962112100190783763061909945889717878397740711340114311597934724670601992737526668932871436226135393872881664511222789565256059138002651403875484920711316522536260604255269532161594824301047729082877262812899724246757871448545439896
p = gmpy2.iroot(n,2)[0]
print(p)
phi = p*(p-1)
d = gmpy2.invert(e,phi)
flag =pow(ct,d,n)
print(bytes.fromhex(hex(flag)[2:]))
print(libnum.n2s(int(flag)))
Everything is Big
低解密指数攻击，这边推文：[(https://blog.csdn.net/weixin_44617902/article/details/113131016)]
解密代码如下：
import gmpy2
from Crypto.Util.number import long_to_bytes
# numerator(n):分子, denominator(d)：分母
def t_cf(n, d): # 将分数 x/y 转为连分数的形式
res = []
while d:
res.append(n // d)
n, d = d, n % d
return res
def cf(sub_res): # 得到渐进分数的分母和分子
n, d = 1, 0
for i in sub_res[::-1]: # 从后面往前循环
d, n = n, i * n + d
return d, n
def list_fraction(x, y): # 列出每个渐进分数
res = t_cf(x, y)
res = list(map(cf, (res[0:i] for i in range(1, len(res))))) # 将连分数的结果逐一截取以求渐进分数
return res
def get_pq(a, b, c): # 由p+q和pq的值通过维达定理来求解p和q(解二元一次方程)
par = gmpy2.isqrt(b * b - 4 * a * c) # 由上述可得，开根号一定是整数，因为有解
x1, x2 = (-b + par) // (2 * a), (-b - par) // (2 * a)
return x1, x2
def wienerAttack(e, n):
for (d, k) in list_fraction(e, n): # 用一个for循环来注意试探e/n的连续函数的渐进分数，直到找到一个满足条件的渐进分数
if k == 0: # 可能会出现连分数的第一个为0的情况，排除
continue
if (e * d - 1) % k != 0: # ed=1 (mod φ(n)) 因此如果找到了d的话，(ed-1)会整除φ(n),也就是存在k使得(e*d-1)//k=φ(n)
continue
phi = (e * d - 1) // k # 这个结果就是 φ(n)
px, qy = get_pq(1, n - phi + 1, n)
if px * qy == n:
p, q = abs(int(px)), abs(int(qy)) # 可能会得到两个负数，负负得正未尝不会出现
d = gmpy2.invert(e, (p - 1) * (q - 1)) # 求ed=1 (mod φ(n))的结果，也就是e关于 φ(n)的乘法逆元d
return d
print("求解d失败")
n = 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
e = 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
c = 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
d = wienerAttack(e, n)
m = pow(c,d,n)
print(long_to_bytes(m))
Crossed Wires
题目：
from Crypto.Util.number import getPrime, long_to_bytes, bytes_to_long, inverse
import math
from gmpy2 import next_prime
FLAG = b"crypto{????????????????????????????????????????????????}"
p = getPrime(1024)
q = getPrime(1024)
N = p*q
phi = (p-1)*(q-1)
e = 0x10001
d = inverse(e, phi)
my_key = (N, d)
friends = 5
friend_keys = [(N, getPrime(17)) for _ in range(friends)]
cipher = bytes_to_long(FLAG)
for key in friend_keys:
cipher = pow(cipher, key[1], key[0])
print(f"My private key: {my_key}")
print(f"My Friend's public keys: {friend_keys}")
print(f"Encrypted flag: {cipher}")
解题代码如下：
from Crypto.Util.number import *
My_private= (21711308225346315542706844618441565741046498277716979943478360598053144971379956916575370343448988601905854572029635846626259487297950305231661109855854947494209135205589258643517961521594924368498672064293208230802441077390193682958095111922082677813175804775628884377724377647428385841831277059274172982280545237765559969228707506857561215268491024097063920337721783673060530181637161577401589126558556182546896783307370517275046522704047385786111489447064794210010802761708615907245523492585896286374996088089317826162798278528296206977900274431829829206103227171839270887476436899494428371323874689055690729986771, 2734411677251148030723138005716109733838866545375527602018255159319631026653190783670493107936401603981429171880504360560494771017246468702902647370954220312452541342858747590576273775107870450853533717116684326976263006435733382045807971890762018747729574021057430331778033982359184838159747331236538501849965329264774927607570410347019418407451937875684373454982306923178403161216817237890962651214718831954215200637651103907209347900857824722653217179548148145687181377220544864521808230122730967452981435355334932104265488075777638608041325256776275200067541533022527964743478554948792578057708522350812154888097)
My_Friend=[(21711308225346315542706844618441565741046498277716979943478360598053144971379956916575370343448988601905854572029635846626259487297950305231661109855854947494209135205589258643517961521594924368498672064293208230802441077390193682958095111922082677813175804775628884377724377647428385841831277059274172982280545237765559969228707506857561215268491024097063920337721783673060530181637161577401589126558556182546896783307370517275046522704047385786111489447064794210010802761708615907245523492585896286374996088089317826162798278528296206977900274431829829206103227171839270887476436899494428371323874689055690729986771, 106979), (21711308225346315542706844618441565741046498277716979943478360598053144971379956916575370343448988601905854572029635846626259487297950305231661109855854947494209135205589258643517961521594924368498672064293208230802441077390193682958095111922082677813175804775628884377724377647428385841831277059274172982280545237765559969228707506857561215268491024097063920337721783673060530181637161577401589126558556182546896783307370517275046522704047385786111489447064794210010802761708615907245523492585896286374996088089317826162798278528296206977900274431829829206103227171839270887476436899494428371323874689055690729986771, 108533), (21711308225346315542706844618441565741046498277716979943478360598053144971379956916575370343448988601905854572029635846626259487297950305231661109855854947494209135205589258643517961521594924368498672064293208230802441077390193682958095111922082677813175804775628884377724377647428385841831277059274172982280545237765559969228707506857561215268491024097063920337721783673060530181637161577401589126558556182546896783307370517275046522704047385786111489447064794210010802761708615907245523492585896286374996088089317826162798278528296206977900274431829829206103227171839270887476436899494428371323874689055690729986771, 69557), (21711308225346315542706844618441565741046498277716979943478360598053144971379956916575370343448988601905854572029635846626259487297950305231661109855854947494209135205589258643517961521594924368498672064293208230802441077390193682958095111922082677813175804775628884377724377647428385841831277059274172982280545237765559969228707506857561215268491024097063920337721783673060530181637161577401589126558556182546896783307370517275046522704047385786111489447064794210010802761708615907245523492585896286374996088089317826162798278528296206977900274431829829206103227171839270887476436899494428371323874689055690729986771, 97117), (21711308225346315542706844618441565741046498277716979943478360598053144971379956916575370343448988601905854572029635846626259487297950305231661109855854947494209135205589258643517961521594924368498672064293208230802441077390193682958095111922082677813175804775628884377724377647428385841831277059274172982280545237765559969228707506857561215268491024097063920337721783673060530181637161577401589126558556182546896783307370517275046522704047385786111489447064794210010802761708615907245523492585896286374996088089317826162798278528296206977900274431829829206103227171839270887476436899494428371323874689055690729986771, 103231)]
c =20304610279578186738172766224224793119885071262464464448863461184092225736054747976985179673905441502689126216282897704508745403799054734121583968853999791604281615154100736259131453424385364324630229671185343778172807262640709301838274824603101692485662726226902121105591137437331463201881264245562214012160875177167442010952439360623396658974413900469093836794752270399520074596329058725874834082188697377597949405779039139194196065364426213208345461407030771089787529200057105746584493554722790592530472869581310117300343461207750821737840042745530876391793484035024644475535353227851321505537398888106855012746117
N = 21711308225346315542706844618441565741046498277716979943478360598053144971379956916575370343448988601905854572029635846626259487297950305231661109855854947494209135205589258643517961521594924368498672064293208230802441077390193682958095111922082677813175804775628884377724377647428385841831277059274172982280545237765559969228707506857561215268491024097063920337721783673060530181637161577401589126558556182546896783307370517275046522704047385786111489447064794210010802761708615907245523492585896286374996088089317826162798278528296206977900274431829829206103227171839270887476436899494428371323874689055690729986771
f_d = []
for i in range(len(My_Friend)):
f_d.append(My_Friend[i][1])
print(f_d)
p = 134460556242811604004061671529264401215233974442536870999694816691450423689575549530215841622090861571494882591368883283016107051686642467260643894947947473532769025695530343815260424314855023688439603651834585971233941772580950216838838690315383700689885536546289584980534945897919914730948196240662991266027
q = 161469718942256895682124261315253003309512855995894840701317251772156087404025170146631429756064534716206164807382734456438092732743677793224010769460318383691408352089793973150914149255603969984103815563896440419666191368964699279209687091969164697704779792586727943470780308857107052647197945528236341228473
phi = (p-1)*(q-1)
for i in f_d:
d = pow(i,-1,phi)
c = pow(c,d,N)
print(c)
print(long_to_bytes(c))
Everything is Still Big
还是和上上道题一样：
#!/usr/bin/env python3
from Crypto.Util.number import getPrime, bytes_to_long, inverse
from random import getrandbits
from math import gcd
FLAG = b"crypto{?????????????????????????????????????}"
m = bytes_to_long(FLAG)
def get_huge_RSA():
p = getPrime(1024)
q = getPrime(1024)
N = p*q
phi = (p-1)*(q-1)
while True:
d = getrandbits(512)
if (3*d)**4 > N and gcd(d,phi) == 1:
e = inverse(d, phi)
break
return N,e
N, e = get_huge_RSA()
c = pow(m, e, N)
print(f'N = {hex(N)}')
print(f'e = {hex(e)}')
print(f'c = {hex(c)}')
解题代码：
import gmpy2
from Crypto.Util.number import long_to_bytes
# numerator(n):分子, denominator(d)：分母
def t_cf(n, d): # 将分数 x/y 转为连分数的形式
res = []
while d:
res.append(n // d)
n, d = d, n % d
return res
def cf(sub_res): # 得到渐进分数的分母和分子
n, d = 1, 0
for i in sub_res[::-1]: # 从后面往前循环
d, n = n, i * n + d
return d, n
def list_fraction(x, y): # 列出每个渐进分数
res = t_cf(x, y)
res = list(map(cf, (res[0:i] for i in range(1, len(res))))) # 将连分数的结果逐一截取以求渐进分数
return res
def get_pq(a, b, c): # 由p+q和pq的值通过维达定理来求解p和q(解二元一次方程)
par = gmpy2.isqrt(b * b - 4 * a * c) # 由上述可得，开根号一定是整数，因为有解
x1, x2 = (-b + par) // (2 * a), (-b - par) // (2 * a)
return x1, x2
def wienerAttack(e, n):
for (d, k) in list_fraction(e, n): # 用一个for循环来注意试探e/n的连续函数的渐进分数，直到找到一个满足条件的渐进分数
if k == 0: # 可能会出现连分数的第一个为0的情况，排除
continue
if (e * d - 1) % k != 0: # ed=1 (mod φ(n)) 因此如果找到了d的话，(ed-1)会整除φ(n),也就是存在k使得(e*d-1)//k=φ(n)
continue
phi = (e * d - 1) // k # 这个结果就是 φ(n)
px, qy = get_pq(1, n - phi + 1, n)
if px * qy == n:
p, q = abs(int(px)), abs(int(qy)) # 可能会得到两个负数，负负得正未尝不会出现
d = gmpy2.invert(e, (p - 1) * (q - 1)) # 求ed=1 (mod φ(n))的结果，也就是e关于 φ(n)的乘法逆元d
return d
print("求解d失败")
n = 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
e = 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
c = 0xa3bce6e2e677d7855a1a7819eb1879779d1e1eefa21a1a6e205c8b46fdc020a2487fdd07dbae99274204fadda2ba69af73627bdddcb2c403118f507bca03cb0bad7a8cd03f70defc31fa904d71230aab98a10e155bf207da1b1cac1503f48cab3758024cc6e62afe99767e9e4c151b75f60d8f7989c152fdf4ff4b95ceed9a7065f38c68dee4dd0da503650d3246d463f504b36e1d6fafabb35d2390ecf0419b2bb67c4c647fb38511b34eb494d9289c872203fa70f4084d2fa2367a63a8881b74cc38730ad7584328de6a7d92e4ca18098a15119baee91237cea24975bdfc19bdbce7c1559899a88125935584cd37c8dd31f3f2b4517eefae84e7e588344fa5
d = wienerAttack(e, n)
m = pow(c,d,n)
print(long_to_bytes(m))
Endless Emails
Infinite Descent
题目代码：
#!/usr/bin/env python3
import random
from Crypto.Util.number import bytes_to_long, isPrime
FLAG = b"crypto{???????????????????}"
def getPrimes(bitsize):
r = random.getrandbits(bitsize)
p, q = r, r
while not isPrime(p):
p += random.getrandbits(bitsize//4)
while not isPrime(q):
q += random.getrandbits(bitsize//8)
return p, q
m = bytes_to_long(FLAG)
p, q = getPrimes(2048)
n = p * q
e = 0x10001
c = pow(m, e, n)
print(f"n = {n}")
print(f"e = {e}")
print(f"c = {c}")
观察到p，q相近，发现能分解n，直接上代码：
from factordb import factordb
n =factordb.FactorDB(383347712330877040452238619329524841763392526146840572232926924642094891453979246383798913394114305368360426867021623649667024217266529000859703542590316063318592391925062014229671423777796679798747131250552455356061834719512365575593221216339005132464338847195248627639623487124025890693416305788160905762011825079336880567461033322240015771102929696350161937950387427696385850443727777996483584464610046380722736790790188061964311222153985614287276995741553706506834906746892708903948496564047090014307484054609862129530262108669567834726352078060081889712109412073731026030466300060341737504223822014714056413752165841749368159510588178604096191956750941078391415634472219765129561622344109769892244712668402761549412177892054051266761597330660545704317210567759828757156904778495608968785747998059857467440128156068391746919684258227682866083662345263659558066864109212457286114506228470930775092735385388316268663664139056183180238043386636254075940621543717531670995823417070666005930452836389812129462051771646048498397195157405386923446893886593048680984896989809135802276892911038588008701926729269812453226891776546037663583893625479252643042517196958990266376741676514631089466493864064316127648074609662749196545969926051)
n.get_factor_list()
n.connect()
n.get_factor_list()
p = n.get_factor_list()[0]
q = n.get_factor_list()[1]
e=65537
c=98280456757136766244944891987028935843441533415613592591358482906016439563076150526116369842213103333480506705993633901994107281890187248495507270868621384652207697607019899166492132408348789252555196428608661320671877412710489782358282011364127799563335562917707783563681920786994453004763755404510541574502176243896756839917991848428091594919111448023948527766368304503100650379914153058191140072528095898576018893829830104362124927140555107994114143042266758709328068902664037870075742542194318059191313468675939426810988239079424823495317464035252325521917592045198152643533223015952702649249494753395100973534541766285551891859649320371178562200252228779395393974169736998523394598517174182142007480526603025578004665936854657294541338697513521007818552254811797566860763442604365744596444735991732790926343720102293453429936734206246109968817158815749927063561835274636195149702317415680401987150336994583752062565237605953153790371155918439941193401473271753038180560129784192800351649724465553733201451581525173536731674524145027931923204961274369826379325051601238308635192540223484055096203293400419816024111797903442864181965959247745006822690967920957905188441550106930799896292835287867403979631824085790047851383294389
import libnum
import gmpy2
from Crypto.Util.number import *
n=p*q
print(n)
fn=(p-1)*(q-1)
print(fn)
d=inverse(e,fn)
flag=int(pow(c,d,n))
print('flag=',flag)
print(libnum.n2s(flag))
print(long_to_bytes(flag))
Marin's Secrets
#!/usr/bin/env python3
import random
from Crypto.Util.number import bytes_to_long, inverse
from secret import secrets, flag
def get_prime(secret):
prime = 1
for _ in range(secret):
prime = prime << 1
return prime - 1
random.shuffle(secrets)
m = bytes_to_long(flag)
p = get_prime(secrets[0])
q = get_prime(secrets[1])
n = p * q
e = 0x10001
c = pow(m, e, n)
print(f"n = {n}")
print(f"e = {e}")
print(f"c = {c}")
p,q相近，和上题一样，解题代码也和上题一样，这里就不赘述了。
Fast Primes
题目：
#!/usr/bin/env python3
import math
import random
from Crypto.Cipher import PKCS1_OAEP
from Crypto.PublicKey import RSA
from Crypto.Util.number import bytes_to_long, inverse
from gmpy2 import is_prime
FLAG = b"crypto{????????????}"
primes = []
def sieve(maximum=10000):
# In general Sieve of Sundaram, produces primes smaller
# than (2*x + 2) for a number given number x. Since
# we want primes smaller than maximum, we reduce maximum to half
# This array is used to separate numbers of the form
# i+j+2ij from others where 1 <= i <= j
marked = [False]*(int(maximum/2)+1)
# Main logic of Sundaram. Mark all numbers which
# do not generate prime number by doing 2*i+1
for i in range(1, int((math.sqrt(maximum)-1)/2)+1):
for j in range(((i*(i+1)) << 1), (int(maximum/2)+1), (2*i+1)):
marked[j] = True
# Since 2 is a prime number
primes.append(2)
# Print other primes. Remaining primes are of the
# form 2*i + 1 such that marked[i] is false.
for i in range(1, int(maximum/2)):
if (marked[i] == False):
primes.append(2*i + 1)
def get_primorial(n):
result = 1
for i in range(n):
result = result * primes[i]
return result
def get_fast_prime():
M = get_primorial(40)
while True:
k = random.randint(2**28, 2**29-1)
a = random.randint(2**20, 2**62-1)
p = k * M + pow(e, a, M)
if is_prime(p):
return p
sieve()
e = 0x10001
m = bytes_to_long(FLAG)
p = get_fast_prime()
q = get_fast_prime()
n = p * q
phi = (p - 1) * (q - 1)
d = inverse(e, phi)
key = RSA.construct((n, e, d))
cipher = PKCS1_OAEP.new(key)
ciphertext = cipher.encrypt(FLAG)
assert cipher.decrypt(ciphertext) == FLAG
exported = key.publickey().export_key()
with open("key.pem", 'wb') as f:
f.write(exported)
with open('ciphertext.txt', 'w') as f:
f.write(ciphertext.hex())
解题代码：
from Crypto.Cipher import PKCS1_OAEP
from Crypto.PublicKey import RSA
from Crypto.Util.number import long_to_bytes, inverse
from gmpy2 import mpz
public = RSA.importKey(open('key.pem').read())
n = int(public.n)
e = int(public.e)
print(n)
print(e)
e=65537
n=4013610727845242593703438523892210066915884608065890652809524328518978287424865087812690502446831525755541263621651398962044653615723751218715649008058509
p=51894141255108267693828471848483688186015845988173648228318286999011443419469
q=77342270837753916396402614215980760127245056504361515489809293852222206596161
assert p*q==n
phi = (p - 1) * (q - 1)
d = inverse(e, phi)
c = open("cipher.txt", 'r').read()
key = RSA.construct((n, e, d))
cipher = PKCS1_OAEP.new(key)
decrypt = cipher.decrypt(bytes.fromhex(c))
print(decrypt)
Ron was Wrong, Whit is Right
题目：
from Crypto.Cipher import PKCS1_OAEP
from Crypto.PublicKey import RSA
msg = "???"
with open('21.pem') as f:
key = RSA.importKey(f.read())
cipher = PKCS1_OAEP.new(key)
ciphertext = cipher.encrypt(msg)
with open('21.ciphertext', 'w') as f:
f.write(ciphertext.hex())
解题代码：
from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_OAEP
from math import gcd
n=[]
c=[]
e=[]
for i in range(1, 51):
key = RSA.importKey(open(f"{i}.pem", 'r').read())
cipher = open(f"{i}.ciphertext", 'r').read()
n.append(key.n)
c.append(cipher)
e.append(key.e)
N = 0
for i in range(len(n)):
for j in range(i,len(n)):
tmp = gcd(n[i], n[j])
if tmp != 1 and n[i]!=n[j]:
N =tmp
index = i
e = e[index]
factor_1 = N
factor_2 = n[index]//N
phi = (factor_1-1)*(factor_2-1)
d = pow(e,-1, phi)
key = RSA.construct((n[index], e, d))
cipher = PKCS1_OAEP.new(key)
decrypt = cipher.decrypt(bytes.fromhex(c[index]))
print(decrypt)
