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ashutosh1206
GitHub Repository: ashutosh1206/crypton
 Path: blob/master/RSA-encryption/Intro-Challenges/Dp&Dq/encrypt.py
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from Crypto.Util.number import *

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import gmpy2

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from Crypto.PublicKey import RSA

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from secret import flag, factor_p, factor_q

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n = 78947048749078921553533397168223306327077414586183944609818553584663665747677757164091069731707888197818123265891690348381736231229561730395657333262905613958377739828118875586000068815867354664296956931106342366972209987312006882239420435532019823504851246162228112328268031305782869824549589268029030234941

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e = 65537

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p = factor_p

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q = factor_q

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assert p*q == n

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phin = (p - 1)*(q - 1)

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d = inverse(e, phin)

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# d congruent to dp (mod p)

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# d congruent to dq (mod q)

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# Apply Chinese Remainder Theorem to solve the equations and get d

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# Ref. to: https://crypto.stanford.edu/pbc/notes/numbertheory/crt.html

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# Ref. to: https://en.wikipedia.org/wiki/Chinese_remainder_theorem

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dp = d % (p-1)

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dq = d % (q-1)

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key = RSA.construct((n, long(e)))

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obj1 = open("publickey.pem",'w')

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obj1.write(key.exportKey('PEM'))

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obj1.close()

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message = bytes_to_long(flag)

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ciphertext = pow(message, e, n)

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ciphertext = long_to_bytes(ciphertext)

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ciphertext = ciphertext.encode("hex")

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f = open("data.txt",'w')

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f.write("[*] ciphertext: " + ciphertext + "\n" + "[*] dp: " + str(dp) + "\n" + "[*] dq: " + str(dq))

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f.close()

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