[HNCTF 2022 WEEK3]Binomials RK1Y8的WriteUp

from Crypto.Util.number import *
from math import gcd

N = 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
e1,e2 = 0x10001,0x65537
c1,c2 = 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,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
c = 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


aa = pow(c1,e2,N)*pow(3,e1*e2,N)%N
bb = pow(c2,e1,N)
q = gcd(aa-bb,N)
p = N//q
assert q*p == N

phin = p*q-p-q+1
d = pow(e1,-1,phin)
print(bytes.fromhex(hex(pow(c,d,N))[2:]))

# b'NSSCTF{M4th_1s_int3res4in9}\x05\x05\x05\x05\x05'
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