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[原创]关于题目more-calc的解法和拓展

发表于: 2020-12-20 16:19 8131

[原创]关于题目more-calc的解法和拓展

happi0

2020-12-20 16:19

8131

看雪似乎不支持数学公式的渲染(还是我没找到?)
数学公式我都用图片替代了,好难受.

由于p高达2048bit,所以极有可能m < p

当m < p 时,下式成立

于是,我们相当与把q给消去了

构造出了新的rsa加密且满足:

然后一个简单的解密即可

且满足∀ m < pi ,(1 <= i <= k) ,则

等效于

其中 i 满足1 <= i <= n

即只需要知道n中的一个素数即可解密

题中首先生成6个素数,并且n由其中三个素数相乘得到

这道题,我们只需要想办法恢复素数集就可以解出了

我们知道每一组的密文和素数在素数集的位置

为了方便描述,我们令p中的素数分别为:p0,p1,p2,p3,p4,p5

于是,我们可以得到如下20个等式:(其实'+'和'-'是一样的,为了好看,统一写作'+')

以第一,第二组方程为例:
可以改写为:

作差可得:

对上式取最大公因式可得p0 * p1

(当然也有可能出现k * pi * pj的情况,不过k通常很小)

组合多组数据再次求公约数就可以解出原本的素数了

实际上,用密文的第一组数据的话,可以只求出前三个素数

由于每一个素数都是1024bit,所以有极大概率m < pi ,(0<= i <= 5)

显然有以下两个式子:

c1−c0=p0∗p1(k1∗p3−k0∗p2)

c7−c0=p0∗(k0∗p1∗p2−k1∗p3∗p4)

两式有公约数p0,可以直接解出p0来作为n

参考1

参考2

import gmpy2

from Crypto.Util.number import *
 
flag = b"flag{xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx}"
 
p = getStrongPrime(2048)
 
for i in range(1, (p+1)//2):

    s += pow(i, p-2, p)
 
s = s % p
 
q = gmpy2.next_prime(s)

n = p*q

e = 0x10001

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

print(p)

print(c)

import gmpy2

from Crypto.Util.number import *

flag = b"flag{xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx}"

p = getStrongPrime(2048)

for i in range(1, (p+1)//2):

s += pow(i, p-2, p)

s = s % p

q = gmpy2.next_prime(s)

n = p*q

e = 0x10001

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

print(p)

print(c)

c = m ^ e mod n

c' = c mod p = m ^ e mod p

c = m ^ e mod n

c' = c mod p = m ^ e mod p

#!/usr/bin/python

import gmpy2

from Crypto.Util.number import long_to_bytes
 
p = 27405107041753266489145388621858169511872996622765267064868542117269875531364939896671662734188734825462948115530667205007939029215517180761866791579330410449202307248373229224662232822180397215721163369151115019770596528704719472424551024516928606584975793350814943997731939996459959720826025110179216477709373849945411483731524831284895024319654509286305913312306154387754998813276562173335189450448233216133842189148761197948559529960144453513191372254902031168755165124218783504740834442379363311489108732216051566953498279198537794620521800773917228002402970358087033504897205021881295154046656335865303621793069

c = 350559186837488832821747843236518135605207376031858002274245004287622649330215113818719954185397072838014144973032329600905419861908678328971318153205085007743269253957395282420325663132161022100365481003745940818974280988045034204540385744572806102552420428326265541925346702843693366991753468220300070888651732502520797002707248604275755144713421649971492440442052470723153111156457558558362147002004646136522011344261017461901953583462467622428810167107079281190209731251995976003352201766861887320739990258601550606005388872967825179626176714503475557883810543445555390014562686801894528311600623156984829864743222963877167099892926717479789226681810584894066635076755996423203380493776130488170859798745677727810528672150350333480506424506676127108526488370011099147698875070043925524217837379654168009179798131378352623177947753192948012574831777413729910050668759007704596447625484384743880766558428224371417726480372362810572395522725083798926133468409600491925317437998458582723897120786458219630275616949619564099733542766297770682044561605344090394777570973725211713076201846942438883897078408067779325471589907041186423781580046903588316958615443196819133852367565049467076710376395085898875495653237178198379421129086523

e = 0x10001

d = gmpy2.invert(e,p-1)

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

print(long_to_bytes(m))

#!/usr/bin/python

import gmpy2

from Crypto.Util.number import long_to_bytes

p =

27405107041753266489145388621858169511872996622765267064868542117269875531364939896671662734188734825462948115530667205007939029215517180761866791579330410449202307248373229224662232822180397215721163369151115019770596528704719472424551024516928606584975793350814943997731939996459959720826025110179216477709373849945411483731524831284895024319654509286305913312306154387754998813276562173335189450448233216133842189148761197948559529960144453513191372254902031168755165124218783504740834442379363311489108732216051566953498279198537794620521800773917228002402970358087033504897205021881295154046656335865303621793069

c =

350559186837488832821747843236518135605207376031858002274245004287622649330215113818719954185397072838014144973032329600905419861908678328971318153205085007743269253957395282420325663132161022100365481003745940818974280988045034204540385744572806102552420428326265541925346702843693366991753468220300070888651732502520797002707248604275755144713421649971492440442052470723153111156457558558362147002004646136522011344261017461901953583462467622428810167107079281190209731251995976003352201766861887320739990258601550606005388872967825179626176714503475557883810543445555390014562686801894528311600623156984829864743222963877167099892926717479789226681810584894066635076755996423203380493776130488170859798745677727810528672150350333480506424506676127108526488370011099147698875070043925524217837379654168009179798131378352623177947753192948012574831777413729910050668759007704596447625484384743880766558428224371417726480372362810572395522725083798926133468409600491925317437998458582723897120786458219630275616949619564099733542766297770682044561605344090394777570973725211713076201846942438883897078408067779325471589907041186423781580046903588316958615443196819133852367565049467076710376395085898875495653237178198379421129086523

e = 0x10001

d = gmpy2.invert(e,p-1)

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

print(long_to_bytes(m))

c = m ^ e mod n

m = c ^ d mod n

c = m ^ e mod n

m = c ^ d mod n

c' = m ^ e mod pi

m = c ^ d mod pi

c' = m ^ e mod pi

m = c ^ d mod pi

from Crypto.Util.number import getPrime

from FLAG import flag
 
flag = int(flag.hex(), 16)

num = 6

e = 0x10001

p = [getPrime(1024) for _ in range(num)]

f = open('cipherlist.txt', 'w')

for i in range(num):

    for j in range(i + 1, num):

        for k in range(j + 1, num):

            n = p[i] * p[j] * p[k]

            f.write(hex(pow(flag, e, n)) + '\n')

from Crypto.Util.number import getPrime

from FLAG import flag

flag = int(flag.hex(), 16)

num = 6

e = 0x10001

p = [getPrime(1024) for _ in range(num)]

f = open('cipherlist.txt', 'w')

for i in range(num):

for j in range(i + 1, num):

for k in range(j + 1, num):

n = p[i] * p[j] * p[k]

f.write(hex(pow(flag, e, n)) + '\n')

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[培训]内核驱动高级班，冲击BAT一流互联网大厂工作，每周日13:00-18:00直播授课

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Editor 雪币： 26245 活跃值： (63297) 能力值： (RANK：135 ) 在线值：发帖 1608 回帖 4397 粉丝 1772 关注私信	Editor 2 楼感谢分享！ 2020-12-20 16:32 0
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