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Last active April 22, 2022 01:29
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def generate_RSA(bits=2048):
'''
Generate an RSA keypair with an exponent of 65537 in PEM format
param: bits The key length in bits
Return private key and public key
'''
from Crypto.PublicKey import RSA
new_key = RSA.generate(bits, e=65537)
public_key = new_key.publickey().exportKey("PEM")
private_key = new_key.exportKey("PEM")
return private_key, public_key
@warsm
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warsm commented May 17, 2017

@miigotu "youthinks" wrong. e should be chosen so that e and λ(n) are coprime. It is not chosen at random, and since it is usually small for computation reasons, and included in the public key, it can always be known by an attacker anyway.

@aravindaran
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from Crypto.PublicKey import RSA
code = 'nooneknows'

key = RSA.generate(2048)
privatekey = key.exportKey(passphrase=code, pkcs=8)
publickey = key.publickey().exportKey()

@WarAtLord
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Nice But How Can I Write The Private Key I Tried This:
f = open('PublicKey.pem','w')
f.write(publick_key)
f.close()

BUT IT DOESN'T WORK WITH THE PRIVATE KEY, JUST RETURNS 0B

@mynameisvinn
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@WarAtLord try publick_key.exportKey("PEM")

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