XCTF分站赛ACTF—

impossible RSA：

没啥好说的，跟我之前文章有道题类似，虽然如此还是花费了很长时间，原因令人落泪，把q = inverse(e,p)的数学式写成了e^q mod p导致数学式推导及其困难（能推但无用）

解题脚本：

#coding:utf-8

from Crypto.Util.number import *

import math

n = 15987576139341888788648863000534417640300610310400667285095951525208145689364599119023071414036901060746667790322978452082156680245315967027826237720608915093109552001033660867808508307569531484090109429319369422352192782126107818889717133951923616077943884651989622345435505428708807799081267551724239052569147921746342232280621533501263115148844736900422712305937266228809533549134349607212400851092005281865296850991469375578815615235030857047620950536534729591359236290249610371406300791107442098796128895918697534590865459421439398361818591924211607651747970679849262467894774012617335352887745475509155575074809

e = 65537

c = 8273086882440893360458062957389163084656045191542493618199369528956277216626884353986044368396198156428766254991928690583227149075264217246716715502497271453823598984519037301602775476502736840821942623288225980044817912940317041496675271105285924648202112216540495276381694590948153181922044287087121526235593090625653756288948499134042427779455887781328892794911088854654421379942237290840799205667104402295294924690771201447934282318850564703279100891083617354084345663030868007048086929831020873706613566948846194280096109248694845560054847526215721665897469865078997234299897107511688667705001432037926136840958

import gmpy2

for k in range(1, 100000000):

    L = gmpy2.iroot(1 + 4 * k * n * e, 2)

    if L[1]:

        p = (-1 + L[0]) // (2 * k)

        q = (p * k + 1) // e

        print(p)

        print(q)

        print(k)

        break

k = 46280

p = 150465840847587996081934790667651610347742504431401795762471467800785876172317705268993152743689967775266712089661128372295606682852482012493939368044600366794969553828079064622047080051569090177885299781981209120854290564064662058027679075401901717932024549311396484660557278975525859127898004619405319768113

q = 106253858346069738600667441477316882476975191191010804704017265511396163224664897689076447029585908855140507431062102645373463498213419404889139172575859514095414665779078979976323891310048026205540865067215318951327289428947198682355325809994354509756230772573224732747769822710641878029801786071777441733193

phi = (p - 1) * (q - 1)

# print(phi)

print(gmpy2.gcd(e, phi))

#ed mod phi 余 1

d = gmpy2.invert(e, phi)

print(d)

m = pow(c, d, n)

# m = m

print(long_to_bytes(m))

flag：ACTF{F1nD1nG_5pEcia1_n_i5_nOt_eA5y}

RSA leak：

题目如下：

from sage.all import *

from secret import flag

from Crypto.Util.number import bytes_to_long

def leak(a, b):

    p = random_prime(pow(2, 64))

    q = random_prime(pow(2, 64))

    n = p*q

    e = 65537

    print(n)

    print((pow(a, e) + pow(b, e) + 0xdeadbeef) % n)

def gen_key():

    a = randrange(0, pow(2,256))

    b = randrange(0, pow(2,256))

    p = pow(a, 4)

    q = pow(b, 4)

    rp = randrange(0, pow(2,24))

    rq = randrange(0, pow(2,24))

    pp = next_prime(p+rp)

    qq = next_prime(q+rq)

    if pp % pow(2, 4) == (pp-p) % pow(2, 4) and qq % pow(2, 4) == (qq-q) % pow(2, 4):

        n = pp*qq

        rp = pp-p

        rq = qq-q

        return n, rp, rq

n, rp, rq = gen_key()

e = 65537

c = pow(bytes_to_long(flag), e, n)

print("n =", n)

print("e =", e)

print("c =", c)

print("=======leak=======")

leak(rp, rq)

'''

n = 3183573836769699313763043722513486503160533089470716348487649113450828830224151824106050562868640291712433283679799855890306945562430572137128269318944453041825476154913676849658599642113896525291798525533722805116041675462675732995881671359593602584751304602244415149859346875340361740775463623467503186824385780851920136368593725535779854726168687179051303851797111239451264183276544616736820298054063232641359775128753071340474714720534858295660426278356630743758247422916519687362426114443660989774519751234591819547129288719863041972824405872212208118093577184659446552017086531002340663509215501866212294702743

e = 65537

c = 48433948078708266558408900822131846839473472350405274958254566291017137879542806238459456400958349315245447486509633749276746053786868315163583443030289607980449076267295483248068122553237802668045588106193692102901936355277693449867608379899254200590252441986645643511838233803828204450622023993363140246583650322952060860867801081687288233255776380790653361695125971596448862744165007007840033270102756536056501059098523990991260352123691349393725158028931174218091973919457078350257978338294099849690514328273829474324145569140386584429042884336459789499705672633475010234403132893629856284982320249119974872840

=======leak=======

122146249659110799196678177080657779971

90846368443479079691227824315092288065

'''

解题思路：

这题我在比赛时也没算出来，在公式的推导过程中就走了弯路导致在有限的时间，有限的算力里面是无法解出答案的。废话不多说，来复盘整理一下思路。

审计代码可以得到如下：

(a^e + b^e + A) mod n1 ≡ B （A,B都是已知的常数，k为未知数，n1为函数里的n，为了与外面n区分写为n1）

p = a⁴ 同理可以得q

pp = rp + p 同理可以得qq

pp mod 16 = pp - p mod 16 同理可以得qq-q

根据同余性质可以得到如下：

(a^e + b^e ) ≡ (B-A) mod n1

a^e ≡ (B-A-b^e ) mod n1

因为leak函数里面又是一个rsa，所以可以求rq和rp，脚本如下：

def get_rq_or_rp():

    #对leak_x分解得到p1,q1

    p = 8949458376079230661

    q = 13648451618657980711

    phi = (p - 1) * (q - 1)

    d = inverse(e, phi)

    for rp in range(10000, pow(2, 24)):

        rq_e = leak_c - 0xdeadbeef   #(a^e + b^e)

        rq_e = (rq_e - pow(rp, e, leak_n)) % leak_n

        rq = pow(rq_e, d, leak_n)

        if len(bin(rq)[2:]) <= 24:

            print(rp)

            print(rq)

            return rp, rq

因为 pp mod 16 = pp-p mod 16，可以推出p+k = pp,因为k远小于p，所以可以近似看成p=pp

则n = pp * qq = p * q

可以推出 pq = (ab)⁴ 从而得到ab，然后得出pq

把 pp = rp + p 以及 qq = rq + q 代入n可以得到如下：

n = (ab)⁴ + a⁴ * rq + b⁴ * rp + rp * rq

可以推出 n - rp * rq - (ab)⁴ = p * rq + q * rp = M

算出M后，两边同乘q，得到如下式子：

rp * q² - M * q + pq * rq = 0

解得q，然后q + rq = qq 同理得 pp

完整脚本如下：

#coding:utf-8

from Crypto.Util.number import *

import gmpy2

leak_n = 122146249659110799196678177080657779971

leak_c = 90846368443479079691227824315092288065

n = 3183573836769699313763043722513486503160533089470716348487649113450828830224151824106050562868640291712433283679799855890306945562430572137128269318944453041825476154913676849658599642113896525291798525533722805116041675462675732995881671359593602584751304602244415149859346875340361740775463623467503186824385780851920136368593725535779854726168687179051303851797111239451264183276544616736820298054063232641359775128753071340474714720534858295660426278356630743758247422916519687362426114443660989774519751234591819547129288719863041972824405872212208118093577184659446552017086531002340663509215501866212294702743

e = 65537

c = 48433948078708266558408900822131846839473472350405274958254566291017137879542806238459456400958349315245447486509633749276746053786868315163583443030289607980449076267295483248068122553237802668045588106193692102901936355277693449867608379899254200590252441986645643511838233803828204450622023993363140246583650322952060860867801081687288233255776380790653361695125971596448862744165007007840033270102756536056501059098523990991260352123691349393725158028931174218091973919457078350257978338294099849690514328273829474324145569140386584429042884336459789499705672633475010234403132893629856284982320249119974872840

def get_rq_or_rp():

    #对leak_x分解得到p1,q1

    p = 8949458376079230661

    q = 13648451618657980711

    phi = (p - 1) * (q - 1)

    d = inverse(e, phi)

    for rp in range(10000, pow(2, 24)):

        rq_e = leak_c - 0xdeadbeef   #(a^e + b^e)

        rq_e = (rq_e - pow(rp, e, leak_n)) % leak_n

        rq = pow(rq_e, d, leak_n)

        if len(bin(rq)[2:]) <= 24:

            return rp, rq

def get_flag(rp, rq):

    #n = pp * qq  因为pp = p + k 又因为k<16所以pp约等于p，同理得q

    pq_ab4 = (gmpy2.iroot(n, 4)[0])**4

    L = n - rp * rq - pq_ab4

    #判别式

    M = gmpy2.iroot(L ** 2 - 4 * rp * rq * pq_ab4, 2)

    if M[1]:

        q1 = (L + M[0]) // (2 * rp)

        q2 = (L - M[0]) // (2 * rp)

        qq1 = q1 + rq

        qq2 = q2 + rq

        pp1 = n // qq1

        pp2 = n // qq2

        if pp1 * qq1 == n:

            phi = (pp1 - 1) * (qq1 - 1)

            d = gmpy2.invert(e, phi)

            m = gmpy2.powmod(c, d, n)

            print(long_to_bytes(m))

        else:

            phi = (pp2 - 1) * (qq2 - 1)

            d = gmpy2.invert(e, phi)

            m = gmpy2.powmod(c, d, n)

            print(long_to_bytes(m))

if __name__ == '__main__':

    rp, rq = get_rq_or_rp()

    if rp and rq:

       get_flag(rp, rq)

flag：ACTF{lsb_attack_in_RSA|a32d7f}

推导过程用到的性质：

同余式相加：若a≡b(mod m)，c≡d(mod m)，则a ± c≡b ± d(mod m)；

不好理解可以如下例子：

17 mod 13 ≡ 4 即 (15 + 2) mod 13 ≡ 4 推出 15 mod 13 ≡ 2

同余其它性质：

传递性：若a≡b(mod m)，b≡c(mod m)，则a≡c(mod m)；

对称性：若a≡b(mod m)，则b≡a (mod m)；

反身性：a≡a (mod m)；

同余式相乘：若a≡b(mod m)，c≡d(mod m)，则ac≡bd(mod m)。

总结：

rsa求解主要通过推导出它们之间的关系，所以想每一题都能做出来，要有一个好的数论基础，没有基础的话就只能像本人一样边做边学，做不做的出来就要靠临场发挥，好的运气（有时候，你推了半天的公式结果根本无法跑出flag，需要的时间太久了）

巴特西

XCTF分站赛ACTF——Crypto

impossible RSA：

RSA leak：

题目如下：

解题思路：

推导过程用到的性质：

总结：

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