Maybe you can use this to do the encoding: hex(rsa.encrypt(4L, '')).rstrip('L').lstrip(0x) You have no method except hacking the source code of the module.įor a long, it returns a long. And you can see from your result of inspect, it only do type conversion to string, no int. So for a number at least 512 int can not satisfy the need. The number is very long, for sha1, it is multiples of 512.

I think the reason to ask a long is, to perform an encryption, you have to do some padding and then perform on the padded number. This will work and the 4L is still a number but it is of type long. If it is a string, it will turns it to long type. The traceback tells clearly the rsa.encrypt wants an argument of long type. Why does it expect the input to be in long format and not int? So, why is it that when I try to encrypt a number using RSA object, it gives an error? If wasString: return tuple(map(long_to_bytes, ciphertext)) Plaintext=bytes_to_long(plaintext) wasString=1 If isinstance(plaintext, types.StringType): '''encrypt(plaintext:string|long, K:string|long) : tupleĮncrypt the string or integer plaintext. Using the inspect module I checked the source code of the encrypt function of RSA object and it says it can encrypt both a string or an integer plaintext: crypto:~$ python > print rsa.encrypt(input,'').encode('hex')įile '/usr/lib/python2.7/dist-packages/Crypto/PublicKey/pubkey.py', line 64, in encryptįile '/usr/lib/python2.7/dist-packages/Crypto/PublicKey/RSA.py', line 71, in _encrypt Type 'help', 'copyright', 'credits' or 'license' for more information. I observed that I can encrypt a string but cannot encrypt an Integer.Ĭould not encrypt the integer, 4: crypto:~$ python

I am trying to encrypt an integer using RSA.

Java rsa biginteger decrypt by python