• 2022鹏城杯CTF---Crypto

    easy_rsa

    加密脚本：

    1. import gmpy2
    2. from Crypto.Util.number import *
    3. import random
    4. from secret import flag
    5.  
    6. m1 = flag[0:12]
    7. m2 = flag[12:24]
    8. m3 = flag[24:]
    9.  
    10. def encrypt1(m):
    11.     while 1:
    12.         e = random.randint(100, 1000)
    13.         p = getPrime(1024)
    14.         q = getPrime(1024)
    15.         phi_n = (p - 1) * (q - 1)
    16.         t = gmpy2.gcd(e, phi_n)
    17.         if gmpy2.invert(e // t, phi_n) and t != 1:
    18.             break
    19.     n = p * q
    20.     c = pow(m, e, n)
    21.     print({'c': format(c, 'x'), 'p': format(p, 'x'), 'q': format(q, 'x'), 'e': format(e, 'x')})
    22.  
    23.  
    24. def encrypt2(m):
    25.     p = getPrime(1024)
    26.     q = getPrime(1024)
    27.     n = p * q
    28.     e = 65537
    29.     c = gmpy2.powmod(m, e, n)
    30.     print({'c': format(c, 'x'), 'p': format((p >> 60) << 60, 'x'), 'e': format(e, 'x'), 'n': format(n, 'x')})
    31.  
    32.  
    33. def encrypt3(m):
    34.     p = getPrime(1024)
    35.     q = getPrime(1024)
    36.     n = p * q
    37.     e = 65537
    38.     M = 2022 * m * 1011 * p
    39.     c = pow(M, e, n)
    40.     print({'c': format(c, 'x'), 'n': format(n, 'x'),'e':format(e, 'x')})
    41.  
    42.  
    43. if __name__ == '__main__':
    44.     encrypt1(bytes_to_long(m1))
    45.     encrypt2(bytes_to_long(m2))
    46.     encrypt3(bytes_to_long(m3))
    47.  
    48. # {'c': '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', 'p': 'bb602e402b68a5cfcc5cfcc63cc82e362e98cb7043817e3421599a4bb8755777c362813742852dad4fec7ec33f1faec04926f0c253f56ab4c4dde6d71627fbc9ef42425b70e5ecd55314e744aa66653103b7d1ba86d1e0e21920a0bfe7d598bd09c3c377a3268928b953005450857c6cfea5bfdd7c16305baed0f0a31ad688bd', 'q': 'bb8d1ea24a3462ae6ec28e79f96a95770d726144afc95ffffa19c7c3a3786a6acc3309820ba7b1a28a4f111082e69e558b27405613e115139b38e799c723ab7fdd7be14b330b118ae60e3b44483a4c94a556e810ab94bbb102286d0100d7c20e7494e20e0c1030e016603bd2a06c1f6e92998ab68e2d420faf47f3ee687fb6d1', 'e': '292'}
    49. # {'c': '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', 'p': 'a9cb9e2eb43f17ad6734356db18ad744600d0c19449fc62b25db7291f24c480217d60a7f87252d890b97a38cc6943740ac344233446eea4084c1ba7ea5b7cf2399d42650b2a3f0302bab81295abfd7cacf248de62d3c63482c5ea8ab6b25cdbebc83eae855c1d07a8cf0408c2b721e43c4ac53262bf9aaf7a000000000000000', 'e': '10001', 'n': '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'}
    50. # {'c': '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', 'n': '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', 'e': '10001'}

    flag被分成了三部分

    第一部分，e与phi_n不互素，gcd(e,phi_n) == 4，此时e对phi_n没有逆元，但gcd(e//t,phi_n) == 1

    故e//t对phi_n有逆元，即m = c ^{t*e/t} mod n，先把c^{t}当整体，再作开方处理得到m。

    1. e1 = 292
    2. p1 = int('bb602e402b68a5cfcc5cfcc63cc82e362e98cb7043817e3421599a4bb8755777c362813742852dad4fec7ec33f1faec04926f0c253f56ab4c4dde6d71627fbc9ef42425b70e5ecd55314e744aa66653103b7d1ba86d1e0e21920a0bfe7d598bd09c3c377a3268928b953005450857c6cfea5bfdd7c16305baed0f0a31ad688bd',16)
    3. q1 = int('bb8d1ea24a3462ae6ec28e79f96a95770d726144afc95ffffa19c7c3a3786a6acc3309820ba7b1a28a4f111082e69e558b27405613e115139b38e799c723ab7fdd7be14b330b118ae60e3b44483a4c94a556e810ab94bbb102286d0100d7c20e7494e20e0c1030e016603bd2a06c1f6e92998ab68e2d420faf47f3ee687fb6d1',16)
    4. c1 = int('27455f081e4858790c6503580dad3302ae359c9fb46dc601eee98f05142128404e95377324720adbbdebf428549008bcd1b670f6749592a171b30316ab707004b9999f3b80de32843afdfd30505b1f4166a03cee9fc48902b74b6e850cfd268e6917c5d84e64f7e7cd0e4a30bfe5903fb5d821d27fdc817d9c4536a8e7aea55af266abcae857a8ffff2f741901baba1b44091a137c69c471c123ab0b80e250e055959c468e6e37c005105ecd7c8b48c659024e5e251df3eeff5da7b3d561cd98150da3575a16bee5f2524d2795fd4879487018345c1e96efc085ed45fb5f02c027aee5bca3aad0eb3e23376c0cd18b02fb05a1ff8fb1af0a3ce4bb671599894e',16)
    5.  
    6. n1 = p1*q1
    7. phi_n = (p1-1)*(q1-1)
    8. t = gcd(e1,phi_n)
    9. d1 = invert(e1//t,phi_n)
    10. m1 = pow(c1,d1,n1)
    11. msg1 = iroot(m1,t)[0]

    第二部分，p高位攻击，利用sage解出p、q来：

    我这里用的在线网站，得到p、q后就能解出m了：

    1. e2 = 10001
    2. n2 = int('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',16)
    3. p2 = 119234372387564173916926418564504307771905987823894721284221707768770334474240277144999791051191061404002537779694672314673997030282474914206610847346023297970473719280866108677835517943804329212840618914863288766846702119011361533150365876285203805100986025166317939702179911918098037294325448226481818486521
    4. q2 = 139862779248852876780236838155351435339041528333485708458669785004897778564234874018135441729896017420539905517964705602836874055417791439544162777504181482765029478481701166935117795286988835104239238153206137155845327225155932803904032184502243017645538314995056944419185855910939481260886933456330514972109
    5. c2 = int('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',16)
    6.  
    7. phi_n2 = (p2-1)*(q2-1)
    8. d2 = invert(e2,phi_n2)
    9. m2 = pow(c2,d2,n2)

    第三部分， 看到M = 2022 * m * 1011 * p，又n = p*q，且c = M^{e} mod n，故求n和c的公因数可以得到p，p出来就好解了：

    1. e3 = 0x10001
    2. n3 = int('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',16)
    3. c3 = int('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',16)
    4.  
    5. p3 = gcd(n3,c3)
    6. q3 = n3 // p3
    7. phi_n3 = (p3-1)*(q3-1)
    8.  
    9. d3 = invert(e3,phi_n3)
    10. M = pow(c3,d3,n3)
    11. m3 = M // (2022 * 1011 * p3)

    注意给的所有数都是16进制的，整体代码如下：

    1. from gmpy2 import *
    2. from Crypto.Util.number import *
    3.  
    4. #part1
    5. e1 = 0x292
    6. p1 = int('bb602e402b68a5cfcc5cfcc63cc82e362e98cb7043817e3421599a4bb8755777c362813742852dad4fec7ec33f1faec04926f0c253f56ab4c4dde6d71627fbc9ef42425b70e5ecd55314e744aa66653103b7d1ba86d1e0e21920a0bfe7d598bd09c3c377a3268928b953005450857c6cfea5bfdd7c16305baed0f0a31ad688bd',16)
    7. q1 = int('bb8d1ea24a3462ae6ec28e79f96a95770d726144afc95ffffa19c7c3a3786a6acc3309820ba7b1a28a4f111082e69e558b27405613e115139b38e799c723ab7fdd7be14b330b118ae60e3b44483a4c94a556e810ab94bbb102286d0100d7c20e7494e20e0c1030e016603bd2a06c1f6e92998ab68e2d420faf47f3ee687fb6d1',16)
    8. c1 = int('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',16)
    9.  
    10. n1 = p1*q1
    11. phi_n = (p1-1)*(q1-1)
    12. t = gcd(e1,phi_n)
    13. d1 = invert(e1//t,phi_n)
    14. m1 = pow(c1,d1,n1)
    15. msg1 = iroot(m1,t)[0]
    16.  
    17. #part2
    18. e2 = 0x10001
    19. n2 = int('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',16)
    20. p2 = 119234372387564173916926418564504307771905987823894721284221707768770334474240277144999791051191061404002537779694672314673997030282474914206610847346023297970473719280866108677835517943804329212840618914863288766846702119011361533150365876285203805100986025166317939702179911918098037294325448226481818486521
    21. q2 = 139862779248852876780236838155351435339041528333485708458669785004897778564234874018135441729896017420539905517964705602836874055417791439544162777504181482765029478481701166935117795286988835104239238153206137155845327225155932803904032184502243017645538314995056944419185855910939481260886933456330514972109
    22. c2 = int('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',16)
    23.  
    24. phi_n2 = (p2-1)*(q2-1)
    25. d2 = invert(e2,phi_n2)
    26. m2 = pow(c2,d2,n2)
    27.  
    28. #part3
    29. e3 = 0x10001
    30. n3 = int('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',16)
    31. c3 = int('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',16)
    32.  
    33. p3 = gcd(n3,c3)
    34. q3 = n3 // p3
    35. phi_n3 = (p3-1)*(q3-1)
    36.  
    37. d3 = invert(e3,phi_n3)
    38. M = pow(c3,d3,n3)
    39. m3 = M // (2022 * 1011 * p3)
    40.  
    41. flag = long_to_bytes(msg1) + long_to_bytes(m2) + long_to_bytes(m3)
    42. print(flag)
    43. #b'PCL{16745c3b0c134c83b74f977260aae9b5}'

    babyrsa

    加密脚本：

    1. from Crypto.Util.number import *
    2. from libnum import s2n
    3. from secret import flag
    4. p = getPrime(1024)
    5. q = getPrime(16)
    6. n = p*q
    7. m = s2n(flag)
    8. for i in range(1,p-q):
    9.     m = m*i%n
    10. e = 1049
    11. print(pow(2,e,n))
    12. print(pow(m,e,n))
    13. #4513855932190587780512692251070948513905472536079140708186519998265613363916408288602023081671609336332823271976169443708346965729874135535872958782973382975364993581165018591335971709648749814573285241290480406050308656233944927823668976933579733318618949138978777831374262042028072274386196484449175052332019377
    14. #3303523331971096467930886326777599963627226774247658707743111351666869650815726173155008595010291772118253071226982001526457616278548388482820628617705073304972902604395335278436888382882457685710065067829657299760804647364231959804889954665450340608878490911738748836150745677968305248021749608323124958372559270

    记两输出结果分别为c1、c2，由print(pow(2,e,n))可以得到：c1 = 2^{e} mod n，进一步得到：

    kn = 2^{e} - c1，n是小于1040位的数，通过计算2^{e}和c1的位数，知道2^{e}是1050位，c1是1039位，

    故可以确定k是小于11位的，这里可以爆破出k来：

    1. k_n = 2**e - c1
    2. for k in range(1,2**11):
    3.     if k_n % k == 0 and (k_n//k).bit_length() <= 1040:#找到位数小于1040的n
    4.         print(k)
    5.         n = k_n // k
    6.         print(n)

    对n进行分解，得到p、q，小的是q。记print(pow(m,e,n))里的m为M，故有c2 = M^{e} mod n，由上面的for循环可以知道M与初始m的关系：M = m * (p-q-1)! mod n，这里需要用威尔逊定理。

    1. p = 170229264879724117919007372149468684565431232721075153274808454126426741324966131188484635914814926870341378228417496808202497615585946352638507704855332363766887139815236730403246238633855524068161116748612090155595549964229654262432946553891601975628848891407847198187453488358420350203927771308228162321231
    2. q =  34211
    3. n = p*q
    4. phi_n = (p-1)*(q-1)
    5. d = invert(e,phi_n)
    6. M = pow(c2,d,n)
    7. for i in range(p-q, p):
    8.     M = M*i%n
    9. m=(-M)%p
    10. print(long_to_bytes(m))

    整体代码如下：

    1. from gmpy2 import *
    2. from Crypto.Util.number import *
    3.  
    4. e = 1049
    5. c1 = 4513855932190587780512692251070948513905472536079140708186519998265613363916408288602023081671609336332823271976169443708346965729874135535872958782973382975364993581165018591335971709648749814573285241290480406050308656233944927823668976933579733318618949138978777831374262042028072274386196484449175052332019377
    6. c2 = 3303523331971096467930886326777599963627226774247658707743111351666869650815726173155008595010291772118253071226982001526457616278548388482820628617705073304972902604395335278436888382882457685710065067829657299760804647364231959804889954665450340608878490911738748836150745677968305248021749608323124958372559270
    7. #print(2**e)
    8. res = 6032057205060440848842124543157735677050252251748505781796615064961622344493727293370973578138265743708225425014400837164813540499979063179105919597766951022193355091707896034850684039059079180396788349106095584290087446076413771468940477241550670753145517602931224392424029547429993824129889235158145614364972941312
    9. #print(res.bit_length())
    10. #print(c1.bit_length())
    11. k_n = 2**e - c1
    12. for k in range(1,2**11):
    13.     if k_n % k == 0 and (k_n//k).bit_length() <= 1040:#找到位数小于1040的n
    14.         print(k)
    15.         n = k_n // k
    16.         print(n)
    17.  
    18. p = 170229264879724117919007372149468684565431232721075153274808454126426741324966131188484635914814926870341378228417496808202497615585946352638507704855332363766887139815236730403246238633855524068161116748612090155595549964229654262432946553891601975628848891407847198187453488358420350203927771308228162321231
    19. q =  34211
    20. n = p*q
    21. phi_n = (p-1)*(q-1)
    22. d = invert(e,phi_n)
    23. M = pow(c2,d,n)
    24. for i in range(p-q, p):
    25.     M = M*i%n
    26. m=(-M)%p
    27. print(long_to_bytes(m))
    28.
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  • 原文地址：https://blog.csdn.net/Luiino/article/details/126918248