一道密码学趣题

　　今天上网安课的时候，HIT-SZ 的同学发我一道 CISCN2020 华南赛区的一道密码学题目，叫做「easyRSA」。这几年名为「easy」的往往不是很 easy，不过本题难度还是很适合国赛选手的。需要一点点思维能力。

　　题目是交互的，源码如下：

#!/usr/bin/python3 -u

import random
from Crypto.Util.number import *
from gmpy2 import *

a = 0x497a81edc78549df
block_len = 64

def gen_prime(bits_len):
    s = random.getrandbits(block_len)
    
    while True:
        s |= 0xe000000000000003
        p = 0
        for _ in range(bits_len // block_len):
            p = (p << block_len) + s
            s = a * s % 2**block_len
        if is_prime(p):
            return p

n = None
e = 0x10001
flag = open("flag.txt", "rb").read()

with open("challenge.py") as f:
    print(f.read())
    
print("Welcome to my Prime Obsession. Tell me what do you want.\n")
while True:
    print("[1] Generate key")
    print("[2] Get Encrypted flag")
    print("[3] Exit")
    opt = int(input(">>> "))
    if opt == 1:
        p = gen_prime(1024)
        q = gen_prime(1024)
        n1 = p * q
        print('pq =',hex(n1))
        c = getRandomInteger(100)
        p = next_prime(p+c)
        q = next_prime(q+c)
        n = p * q
        print('n =',hex(n)) 
    if opt == 2:
        if not n:
            print("No key generated :(")
        else:
            print('cipher =',hex(pow(bytes_to_long(flag), e, n)))
    if opt == 3:
        print("Good bye! Good luck in this game :)")
        break
    print("\n")

　　首先我们需要注意到两件事：

质数生成是自己实现的。这里可能有漏洞点。
每次交互，一共只提供了 $p\times q$ 以及 $p ^ \prime \times q ^ \prime$ 的值。

　　先考虑第二件事。交互题目可能是从多次交互中获取信息，把这些信息综合起来解题；也可能是做题的成功率并非 100%，需要爆破。前者的典型例子是「爆破梅森旋转」的题目，需要用连续的随机数生成结果来恢复 MT19937 的状态；后者的典型例子是各种有特殊条件要求的数论题目。确定题目到底属于哪一种，可以节省大量的做题时间。

　　本题源码，每次 $p, q$ 是随机生成的，而源码里面密码学不安全的 PRNG，只有用于生成 s 的 random.getrandbits() 。由于后文 if is_prime(p) 的条件极为苛刻，几乎不可能一次就成功，故我们无法拿到当初的 s 。所以本题不是 MT19937 爆破，而是概率性解题。

　　现在来考虑本题的质数生成算法。

def gen_prime(bits_len):
    s = random.getrandbits(block_len)
    
    while True:
        s |= 0xe000000000000003
        p = 0
        for _ in range(bits_len // block_len):
            p = (p << block_len) + s
            s = a * s % 2**block_len
        if is_prime(p):
            return p

　　不难分析出，算法会尝试以下步骤：

$s$ 是一个 64 位数。
把 $s$ 作为一个块，接在目前的结果后面。
$s$ 置为 $a \cdot s$，保留低 64 位。
重复步骤 2、3，直到生成一个 1024 位的数。
如果上述算法成功地生成了一个质数，则返回；否则重复上述算法，直到产出质数为止。

　　我们只关注 s |= 0xe000000000000003 之后的 $s$，称为其所产生的质数的种子。随便模拟一下上面的算法，可以发现：$$p = s \| s\cdot a \| s\cdot a ^ 2 \| \cdots \| s \cdot a ^ {15}$$其中每个块都要模 $2 ^ {64}$。来用代码验证一下：

def test_get_prime(bits_len):
    s = random.getrandbits(block_len)
    
    while True:
        s |= 0xe000000000000003
        
        origin_s = s
        
        p = 0
        for _ in range(bits_len // block_len):
            p = (p << block_len) + s
            s = a * s % 2**block_len
        if is_prime(p):
            return origin_s, p
        
test_get_prime(1024)

# p = 0xe402458e1aa2fbbf171cf392f0fdc2616d735e74acdefb7f7b1d85d98def4aa1c61b9a5dcc5eeb3fa919b57d487ee2e1a59c89641b1ecaff2cf2adce17308b217fd4620efa1a9abfa8cbd04d5188436164d10992494e5a7fce77d1f5f00a0ba1047f6ffe27b60a3f59926694cc39e3e1bb7d3c5eb34da9ffdef8fe30e09bcc21
# s = 0xe402458e1aa2fbbf
# s * a % 2^64 = 0x171cf392f0fdc261
# ...
# s * a^15 % 2^64 = 0xdef8fe30e09bcc21

　　从而，$p$ 的最初 64 位就是它的种子 $s$；$p$ 的末尾 64 位除以 $a ^ {15}$，也可以得到 $s$（运算需要在模 $2 ^ {64}$ 下进行）。如果我们能知道种子，我们就可以直接调用上面的生成算法来产出对应的质数 $p$。于是，种子和质数是一一对应的。

　　但我们现在只知道 $p\times q$，不知道 $p, q$ 各自的种子 $s_1, s_2$. 这里我们注意到两点性质：

$p\times q$ 的低 64 位除以 $a ^ {30}$，一定等于 $s_1 \times s_2$ 的低 64 位。这很容易证明。
$p\times q$ 的高 64 位，可能等于 $s_1 \times s_2$ 的高 64 位，也可能等于这个值加上 1。

　　来验证一下：

s1, p = test_get_prime(1024)
s2, q = test_get_prime(1024)

s1, s2, p, q
'''
(16243107326219462111,
 17432706745670175347,
 158294181414880499770651112900548083954434772914472730549340087434970226774360616234703762340566481057536952421125964808455071783034621762005413503653274430171642354361906447116811174866340950519316076420777049700943049933805894766346310426693961910435971922759361714776387154457211555746264926998149652735489,
 169887201305206840068646556183834380043094239543227647711110290353531119335337866448387930053370040451159123818495584829213442749847091386275830475868414884168339924826011599148098370699032169016502100581570327973098847038588859614367540256927662605166485372884790453289507254371825313084721972174725171008493)
'''

R = Zmod(2 ^ 64)
(p * q) / R(a) ^ 30, R(s1 * s2)
# (8570911218002475309, 8570911218002475309)

((p * q) >> (2048 - 64)), (s1 * s2) >> 64
# (15350206276238985089, 15350206276238985088)

　　于是，我们只要手上有 $p \times q$，我们就可以得到 $s_1 \times s_2$ 的高 64 位和低 64 位（其中高位有两种可能性）。而 $s_1 \times s_2$ 一共才 128 位，我们可以完全恢复出 $s_1 \times s_2$，然后质因数分解之，得到 $s_1, s_2$ 的确切值，从而求出 $p, q$！

　　这里有一个细节：质因数分解 $s_1 \times s_2$ 可能会得到多个质因子，需要找出合适的 $s_1, s_2$. 做法是枚举分解结果的子集，求个乘积，我们已经知道 $s$ 必然落在 $[\text{0xe000000000000003}, 2^{64})$ 内，故很容易判断合法性。

import random
from Crypto.Util.number import *
from gmpy2 import *
import itertools
import functools

a = 0x497a81edc78549df

# 从种子生成质数
def build_p(s):
    assert s >= 0xe000000000000003

    p = 0
    for _ in range(1024 // 64):
        p = (p << 64) + s
        s = a * s % 2**64
    assert is_prime(p)
    return p

assert build_p(17361778395699355931) == 169195982262430730898037798646562657888768606161540984197124026057984801045528687609713183918888150519974429332633850456287413410292677840077409039763029168042539147308080880278827892817507057470081212780508473831501271300644154803233797207298692699435252423812271233139678356819016235706919593528823372354117


# 给定 s1*s2，找出合法的 s1, s2.
def test_s1s2(m):
    r = list(factor(m))
    lis = []
    
    for x, cnt in r:
        for _ in range(cnt):
            lis.append(x)
    
    print(f'factor: {lis}')
    
    ans = []
    
    for num in range(len(lis)):
        for f in itertools.combinations(lis, num):
            if f:
                maybe_s = functools.reduce(lambda x,y: x*y, f)
                
                if 0xe000000000000003 <= maybe_s <= 2**64:
                    ans.append(maybe_s)
    
    return ans
    
test_s1s2(309177609417944057766013526324891519733)
'''
factor: [3, 2539, 8750453, 824716561, 5624563537715845553]
[16873690613147536659, 18323057860089065687]
'''

　　从 $p\times q$ 中恢复 $s_1, s_2$，进而获取 $p, q$ 的代码如下：

# 给定 p*q，恢复出 p, q
def attack_pq(n):
    R = Zmod(2^64)
    
    last = R(n) / R(a^30)
    first = n >> (2048 - 64)
    
    s = test_s1s2(int(int(first-1) << 64) + int(last))
    
    if len(s) != 2 or build_p(s[0]) * build_p(s[1]) != n:
        s = test_s1s2(int(int(first) << 64) + int(last))
    
    print(f's: {s}')
    
    assert s
    p, q = build_p(s[0]), build_p(s[1])
    assert p * q == n
    
    
    print(f'✅ p = {p}')
    print(f'✅ q = {q}')
    
    return p, q

attack_pq(25625978037564023374054812252766044742673016474380449671434417557988683145342557433392338411568097592174126890010367975804729035315018122328818953482824183616532541483357942995500254875116460438812227079806451323262767112848392097604212800284648459461746966222062286184693080278059901247068922366712016948770993253748320480569966269448769253837874539131187463351933833504083780041515646366807785588792568168133501958654184904533902107193125303230406615614009544017592608298466267250350122225533985589679020422790330076477130447738445165355516162779670614159780415964621699239667668730434728470462911970718793385464741)
'''
factor: [7, 73, 263, 1795877, 215313151, 1049925013, 4945460059]
s: [16502574388980455299, 16350727310827583863]
✅ p = 160822769417102009618567552156182921546741447360725683118172696897367965121337165186908622589209139632843608852568696104089644870013508160510721639945162275182824416298942113379063991609668151500017228073190018871134644149081490901642408360701118595923659997953670380025262649577673799718463762187171438880989
✅ q = 159342971958788680920230866791247521236447945260297678924204288359901348836858194337996522293090932934733371676245041069970267950853712578389627531630378737762022350847049839429987844064870186209635909347054818141164832660432576265456168359085362019835152391005771348487649822270428353946391414002824201438569
'''

　　现在我们已经搞到 $p, q$ 了！下一个问题是：RSA 采用的质数是 $p ^ \prime, q ^ \prime$，如何从 $p, q$ 找到它们呢？

　　来分析题目源码，生成 $p^\prime, q ^\prime$ 的算法如下：

p = gen_prime(1024)
q = gen_prime(1024)
n1 = p * q
print('pq =',hex(n1))
c = getRandomInteger(100)
p = next_prime(p+c)
q = next_prime(q+c)
n = p * q
print('n =',hex(n))

　　可见，程序先生成了一对 $p, q$（我们可以求出它们），然后随机生成了一个 100 位的整数 $c$，求出 $p+c, q+c$ 的下一个质数，作为 $p ^ \prime$ 和 $q ^ \prime$.

　　由于「下一个质数」这个函数性质极其烂，相当难以研究，我们直接绕过它，记 $\newcommand\pp{p ^ \prime} \newcommand\qq{q ^ \prime} c_1 := \pp - p, c_2 := \qq - q$，于是可以重新表示 $\pp, \qq$：$$\pp = p + c_1, \qquad \qq = q + c_2$$于是$$\pp \times \qq = (p + c_1)(q + c_2) = pq + c_1 q + c_2 p + c_1 c_2$$上面的式子中，$p\times q$ 和 $\pp\times \qq$ 是我们已知的，于是可以求出 $$\pp\times\qq - p\times q = c_1 q + c_2 p + c_1 c_2$$我们把这个值记为 $\Delta$，它是已知量。接下来要干的事情是解出这个不定方程，拿到 $c_1, c_2$，然后求出 $\pp, \qq$.

　　事实上，上面的话说起来容易，做起来难。$c$ 有 100 位，$c_1, c_2$ 比 $c$ 还大一点点，所以是不可枚举的；这个方程有俩未知数，还是二次的，解起来相当困难。

　　开点脑洞：如果 $c_1 = c_2$，会是什么情形？整个方程变成了一元二次方程，可以快速求解了：$$\Delta = x^2 + px + qx$$

　　那么，$c_1 = c_2$ 的概率是多少呢？我们考虑 $c_1$ 与 $c$ 的距离，它等于 $p+c$ 的下一个质数与 $p+c$ 的距离。而由质数分布定理，数量级为 $n$ 的俩质数之间的距离，是 $\ln n$ 量级的。所以 $c_1$ 与 $c$ 的距离，平均来看不会超过 $\ln 2 ^ {1024} \approx 710$ 这个级别；$c_2$ 同理。这样的话，我们预期 $c_1, c_2$ 都在 $[c, c+710]$ 以内取值，它们恰好相等的概率是 $1/710$.

　　由于质数定理的 bound 在 $n$ 很小的时候极其粗糙，故上面的概率计算误差非常大；但至少数量级是没有问题的。所以解题的总体思路就形成了：

疯狂交互，获取海量的 $p\times q, \pp\times\qq$ 组。
假设 $c_1 = c_2$，求出 $\pp$ 和 $\qq$。这大概有 $1/710$ 级别的概率成功。

　　我们交互几千次，事情基本上必定成功了。代码如下：

# 从 p*q, p'*q' 拿到 p', q' 和私钥 d
def work(pq, n):
    try:
        p, q = attack_pq(pq)
        dt = n - pq
#         dt = x(p+q)+x^2

        print(f'dt: {dt}')

        x = var('x')
        ans = solve(x*(p+q) + x^2 == dt, x)
        x1, x2 = ans[0].right(), ans[1].right()
        
        c = x1
        
        if not c.is_integer():
            c = x2
        elif not c.is_positive():
            c = x2
        
        assert c.is_integer()
        assert c.is_positive()
        
        print(f'✅ c = {c}')
        
        
        pp = p + c
        qq = q + c
        
        print(f'✅ p\' = {pp}')
        print(f'✅ q\' = {qq}')

        
        
        assert pp * qq == n
        
        print(f'✅ d = {inverse_mod(0x10001, (pp-1) * (qq-1))}')
        
    except Exception:
        print(f'这组不太行，换一个吧')

work(25625978037564023374054812252766044742673016474380449671434417557988683145342557433392338411568097592174126890010367975804729035315018122328818953482824183616532541483357942995500254875116460438812227079806451323262767112848392097604212800284648459461746966222062286184693080278059901247068922366712016948770993253748320480569966269448769253837874539131187463351933833504083780041515646366807785588792568168133501958654184904533902107193125303230406615614009544017592608298466267250350122225533985589679020422790330076477130447738445165355516162779670614159780415964621699239667668730434728470462911970718793385464741, 28021904302129336729440775301567915332437216258089567111767564794016674457018155360039100520951355548380576284595610697291190503936385180226538571965741142420627721060581430072608205374836150303308140153045011513224970372092841229476265755575324796173767510268509134100146537708142088453311940311115215141352187534401616817279239626963929615354185356780526239992752179350385945092938270110754347951812524816974827773746044388551113627524532864731322649347204697974569586931351733171516384937029102764549501466259800373343884249624949732899362023883491675849231649767788103527947639812648454167039111814437985284546309)

'''
factor: [7, 73, 263, 1795877, 215313151, 1049925013, 4945460059]
s: [16502574388980455299, 16350727310827583863]
✅ p = 160822769417102009618567552156182921546741447360725683118172696897367965121337165186908622589209139632843608852568696104089644870013508160510721639945162275182824416298942113379063991609668151500017228073190018871134644149081490901642408360701118595923659997953670380025262649577673799718463762187171438880989
✅ q = 159342971958788680920230866791247521236447945260297678924204288359901348836858194337996522293090932934733371676245041069970267950853712578389627531630378737762022350847049839429987844064870186209635909347054818141164832660432576265456168359085362019835152391005771348487649822270428353946391414002824201438569
dt: 2395926264565313355385963048801870589764199783709117440333147236027991311675597926646762109383257956206449394585242721486461468621367057897719618482916958804095179577223487077107950499719689864495913073238560189962203259244449131872052955290676336712020544046446847915453457430082187206243017944403198192581194280653296336709273357515160361516310817649338776640818345846302165051422623743946562363019956648841325815091859484017211520331407561500916033733195153956976978632885465921166262711495117174870481043469470296866753801886504567543845861103821061689451233803166404288279971082213725696576199843719191899081568
这组不太行，换一个吧
'''

　　我们来演示一下最终成功的情形：

work(26069239126763544144313564816903584919225468351493641473440054692950282153197487195135311414223645356673159563866063143967003230219599632002379909718878017334229566083385213429707426557827637628431984902602452592854163064832279957432054115424063715513512846471585266280161741825703159883696402247451433556277788724769658068000934928422898992126447860516665865835501335448925851168676214586852842707909375735783055136350498143128185177506086227282670017144764020593609782382354586453914439607583463700059239194111980984987953393701453010426104042888563904056559044144190741477237154627975857816010726687018495994122029, 26069239126763544144313564816903584919225468351493641473440054692950282153197487195135311414223645356673159563866063143967003230219599632002379909718878017334229566083385213429707426557827637628431984902602452592854163064832279957432054115424063715513512846471585266280161741825988418992958523507177816456088510986107884156402419647457685387283014663915814410392002684523734647052244297209296553241947696498795250861154920248758349455971520660721702370823864378454485566700875895894094536122947448903755745971852539042819192334317553458436695944239919353727836898891349048982180198805160517418889084394272238730812629)

'''
factor: [5, 11, 331, 33132481, 9291179069, 48980273938841381]
s: [16212470673756497111, 16931189768418210395]
✅ p = 157995617616370774660638102567731907466273813434698819562579909149220302616489678170556176334851823929143192096314898180613977031807508943016229435352121365585415987271997059622520528656905503238568143547394575088942893875103855187924344429830281996743274207759666120554509518173226980108459374040551600080393
✅ q = 164999760879838291412777527823360433620664042304532719501056970192987047135901399271919480366966970340411907976436953044179019339177047206138418296577749301575525466287977153837203774422689670859379538394344219536816095808445638679593826944251873998482593615255116197397498058399989519359456583919883121928453
dt: 285259109262121259726382899810722261338226088401484719034786395156566803399148544556501349074808795883568082622443710534038320763012195724804422105630164278465434433439032353679100357860875784318521309440180096515363985203696506777740558057831238940616100448010591901351355449671277854747158307504943044177184659602878357707253742736690600
✅ c = 883167773453047379752624032350
✅ p' = 157995617616370774660638102567731907466273813434698819562579909149220302616489678170556176334851823929143192096314898180613977031807508943016229435352121365585415987271997059622520528656905503238568143547394575088942893875103855187924344429830281996743274207759666120554509518174110147881912421420304224112743
✅ q' = 164999760879838291412777527823360433620664042304532719501056970192987047135901399271919480366966970340411907976436953044179019339177047206138418296577749301575525466287977153837203774422689670859379538394344219536816095808445638679593826944251873998482593615255116197397498058400872687132909631299635745960803
✅ d = 18929902510086673211830542076561708694646090199112423578734741638811831139486324032673060025504509801771808146311580923115884412208684054615885034768324548986353522748099835541851880959770197706630601485114486724160958940633586689873415372978863362033241723770332350229742239219936263049207366452280163988720581935255212216393630786066330594961493485442924114415254227968155268214930303811721679329298962022684488694541088421353196184688495242084669325636237107146970285348650646151940584092889379909679798630064926270828345811339970011899882524877738169322630366930298295867738153903553061710361967599556469562610621
'''

　　至此问题解决。本题目一共两个大的难点：一是找到质数生成算法的漏洞；二是从 $p,q$ 出发拿到 $\pp, \qq$. 确实是很有价值的题目，值得密码学选手做一做。

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