pr
from Crypto.Util.number import *
import random
flag=plaintext = 'NSSCTF{****************}'
charset = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'
padding_length = 100 - len(plaintext)
for _ in range(padding_length):
plaintext += random.choice(charset)
public_exponent = 31413537523
message = bytes_to_long(plaintext.encode())
assert message > (1 << 512)
assert message < (1 << 1024)
prime_p = getPrime(512)
prime_q = getPrime(512)
prime_r = getPrime(512)
n1 = prime_p * prime_q
n2 = prime_q * prime_r
ciphertext1 = pow(message, public_exponent, n1)
ciphertext2 = pow(message, public_exponent, n2)
print('c1=', ciphertext1)
print('c2=', ciphertext2)
print('p=', prime_p)
print('r=', prime_r)
'''
c1= 36918910341116680090654563538246204134840776220077189276689868322808977412566781872132517635399441578464309667998925236488280867210758507758915311644529399878185776345227817559234605958783077866016808605942558810445187434690812992072238407431218047312484354859724174751718700409405142819140636116559320641695
c2= 15601788304485903964195122196382181273808496834343051747331984997977255326224514191280515875796224074672957848566506948553165091090701291545031857563686815297483181025074113978465751897596411324331847008870832527695258040104858667684793196948970048750296571273364559767074262996595282324974180754813257013752
p= 12101696894052331138951718202838643670037274599483776996203693662637821825873973767235442427190607145999472731101517998719984942030184683388441121181962123
r= 10199001137987151966640837133782537428248507382360655526592866939552984259171772190788036403425837649697437126360866173688083643144865107648483668545682383
'''
|
分析题目，生成了512位大素数p,q,r
密文如下：
c1 = m ** e (mod n1)
c2 = m ** e (mod n2)
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附件数据给了密文c1,c2和p,r，求m
由于n1和n2有公因子q，根据同余的性质，可以转换为以下形式(此处用到性质3.3)
c1 = m ** e (mod p)
c2 = m ** e (mod r)
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附：[同余的性质](同余基本性质_百度百科 (baidu.com))，内容来自百度百科
根据题目的CRT就可以得到：
然后由于flag只填充到100字节，所以就800bit左右，在模pr下一定能解出明文
exp:
from sympy.ntheory.modular import crt
from Crypto.Util.number import *
c1= 36918910341116680090654563538246204134840776220077189276689868322808977412566781872132517635399441578464309667998925236488280867210758507758915311644529399878185776345227817559234605958783077866016808605942558810445187434690812992072238407431218047312484354859724174751718700409405142819140636116559320641695
c2= 15601788304485903964195122196382181273808496834343051747331984997977255326224514191280515875796224074672957848566506948553165091090701291545031857563686815297483181025074113978465751897596411324331847008870832527695258040104858667684793196948970048750296571273364559767074262996595282324974180754813257013752
p= 12101696894052331138951718202838643670037274599483776996203693662637821825873973767235442427190607145999472731101517998719984942030184683388441121181962123
r= 10199001137987151966640837133782537428248507382360655526592866939552984259171772190788036403425837649697437126360866173688083643144865107648483668545682383
e = 31413537523
n = [p,r]
c = [c1,c2]
M = crt(n,c)[0]
phi = (p-1)*(r-1)
d = inverse(e,phi)
print(long_to_bytes(pow(M,d,p*r)))
|
break
パスワード(文件名)
6081370370545409218106271903400346695565292992689150366474451604281551878507114813906275593034729563149286993189430514737137534129570304832172520820901940874698337733991868650159489601159238582002010625666203730677577976307606665760650563172302688129824842780090723167480409842707790983962415315804311334507726664838464859751689906850572044873633896253285381878416855505301919877714965930289139921111644393144686543207867970807469735534838601255712764863973853116693691206791007433101433703535127367245739289103650669095061417223994665200039533840922696282929063608853551346533188464573323230476645532002621795338655
|
pri-break.pem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-----END PRIVATE KEY-----
|
题目给了私钥的一部分，之前刚好看过”手撕私钥”，这里可以参考这篇文章：
手撕PEM密钥（RSA） | Tover’ Blog
那么先把内容转为十六进制
import libnum
import base64
with open('./pri-break.pem', 'r') as f:
data = f.read()
key_64 = ''.join(data.split('\n')[0:-1])
key_num = libnum.s2n(base64.b64decode(key_64))
key_hex = hex(key_num)[2:]
print(key_hex)
'''
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
'''
|
下表是常见的证书参数
|
|
|
|
| 数据 |
长度 |
名称 |
含义 |
| 30 |
1 |
标记符 |
代表ASN.1结构的开始 |
| 82 |
1 |
长度类型 |
代表后面跟着一个双字节长度 |
| 025b |
2 |
长度 |
代表后续内容的总长度为603字节 |
| 02 |
1 |
类型 |
代表整型 |
| 01 |
1 |
长度 |
代表1字节 |
| 00 |
1 |
值 |
代表整数0 |
| 02 |
1 |
类型 |
代表整型 |
| 81 |
1 |
长度类型 |
代表后面跟着一个单字节长度 |
| 80 |
1 |
长度 |
代表数据长度为128字节 |
| … |
128 |
值 |
数据值1 |
| … |
|
|
|
对信息分段后大概可以得到：
5cf2d493aef1899b6a18b82cc8cfe620e27c3
q (028181)= 00e26146b07b963eacf6c585538c541f190727cc6b82e37920dd676de3106bf29411d27b2bc5090ab7212504e349350f65e699e69930bddad67527e8e448586686da985cc2b58b911ea7b9b5666f54094b38339851c69cbe7a2870726710fdaba0cf07ea3a8065adf3fe9d741976348654ff56ed74f420a237c05d7245cd1f7645
dp (028181)= 00bbaa507453b9b653815d1bd79e95277e59228d515965640b878e2ee3821ecec7a2daf713799a2199d72bcfe38c97db5f0e27d662d3ca3abc8acce848a5e392b34ed1a6e22b889fc08ca68c7ee5e52da31cafc5af6bb365aef8845937dcc2304fa9faa23cfd31b6edd2320f3b1d80da9c528f044a5b3bb96e7f14403b6d84c289
dq (028181)= 00bbd4ac9b781fa4674ce57c8762f2d54ea5e23eb9a1c36dd877fbfe701d4b03de199f83b5406a0d2aff6440d6f901fe81b54152d51e5d18bb423fd7fbb98f279a92bb429407a002cbe5acafb78b7db5ac64b86294cf6f7497d4ffdfdc667ba3b0ecb68b80f8d1e4f082fd4fbc64989630e61fd12c6df6b9c1fa694e8c5df2d1e5
q^(-1) mod p (028180)= 3fec91f745c465b184793a535992dbb20cf6324d83dc4bcf99f1ce4a8f3e12e7fab760208b93b90b6ffbe7848c43fefbb655d74ae0e7229525dfdbc72f07d8837b690a3b9195ae9474d00d9305c203a19f4bba825262faa0bc6d6cd80c5390485ad8f57437bf028cb29aa6fbaf7bc4cf9d3145b42cc2faa1413d663122018ff9
|
flag较短的话就可以直接在模q下解密
exp:
import gmpy2
q = 0x00e26146b07b963eacf6c585538c541f190727cc6b82e37920dd676de3106bf29411d27b2bc5090ab7212504e349350f65e699e69930bddad67527e8e448586686da985cc2b58b911ea7b9b5666f54094b38339851c69cbe7a2870726710fdaba0cf07ea3a8065adf3fe9d741976348654ff56ed74f420a237c05d7245cd1f7645
dp = 0x00bbaa507453b9b653815d1bd79e95277e59228d515965640b878e2ee3821ecec7a2daf713799a2199d72bcfe38c97db5f0e27d662d3ca3abc8acce848a5e392b34ed1a6e22b889fc08ca68c7ee5e52da31cafc5af6bb365aef8845937dcc2304fa9faa23cfd31b6edd2320f3b1d80da9c528f044a5b3bb96e7f14403b6d84c289
dq = 0x00bbd4ac9b781fa4674ce57c8762f2d54ea5e23eb9a1c36dd877fbfe701d4b03de199f83b5406a0d2aff6440d6f901fe81b54152d51e5d18bb423fd7fbb98f279a92bb429407a002cbe5acafb78b7db5ac64b86294cf6f7497d4ffdfdc667ba3b0ecb68b80f8d1e4f082fd4fbc64989630e61fd12c6df6b9c1fa694e8c5df2d1e5
n = 0x3fec91f745c465b184793a535992dbb20cf6324d83dc4bcf99f1ce4a8f3e12e7fab760208b93b90b6ffbe7848c43fefbb655d74ae0e7229525dfdbc72f07d8837b690a3b9195ae9474d00d9305c203a19f4bba825262faa0bc6d6cd80c5390485ad8f57437bf028cb29aa6fbaf7bc4cf9d3145b42cc2faa1413d663122018ff9
c = 6081370370545409218106271903400346695565292992689150366474451604281551878507114813906275593034729563149286993189430514737137534129570304832172520820901940874698337733991868650159489601159238582002010625666203730677577976307606665760650563172302688129824842780090723167480409842707790983962415315804311334507726664838464859751689906850572044873633896253285381878416855505301919877714965930289139921111644393144686543207867970807469735534838601255712764863973853116693691206791007433101433703535127367245739289103650669095061417223994665200039533840922696282929063608853551346533188464573323230476645532002621795338655
p = n//q
m = pow(c,dq,q)
print('-------------------')
print(m)
print(hex(m)[2:])
print(bytes.fromhex(hex(m)[2:]))
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