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Home/ Questions/Q 6388561
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T03:17:39+00:00 2026-05-25T03:17:39+00:00

I’m using the RSA implementation in PyCrypto. With regard to the encrypt(self, plaintext, K)

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I’m using the RSA implementation in PyCrypto. With regard to the encrypt(self, plaintext, K) method K is a parameter of random data. I want to know how much random data needs to be passed in order for the encryted data to be considered secure. For example in my implementation I am passing a strong prime number of 1024 bits via the Crypto.Util.number module like so:

enc_data = public_key.encrypt(data, number.getPrime(1024))

Is this considered ‘secure enough’?

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  1. Editorial Team
    Editorial Team
    2026-05-25T03:17:40+00:00Added an answer on May 25, 2026 at 3:17 am

    The RSA implementation does not use the K parameter. You may ignore it; the RSA implemention does.

    Looking at lines 59-60 of pycrypto-2.3/lib/Crypto/PublicKey/RSA.py you see the following:

    def _encrypt(self, c, K):
        return (self.key._encrypt(c),)
    

    Which proves that K, if supplied, is ignored.

    Official documentation

    Plus, the developers declare this explicitly in the documentation. In fact, if you create a public key public_key and you type

    help(public_key.encrypt)
    

    you will obtain their documentation, which explicitly says:

    encrypt(self, plaintext, K) method of Crypto.PublicKey.RSA._RSAobj instance
    Encrypt a piece of data with RSA.
    
    ...
    ...
    
    :Parameter K: A random parameter (*for compatibility only. This
     value will be ignored*)
    :Type K: byte string or long
    
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