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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 |
e should be random methinks =P
@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.
from Crypto.PublicKey import RSA
code = 'nooneknows'
key = RSA.generate(2048)
privatekey = key.exportKey(passphrase=code, pkcs=8)
publickey = key.publickey().exportKey()
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
@WarAtLord try publick_key.exportKey("PEM")
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Gives me error: