[pycrypto] PyCrypto AND Crypt_RSA integration

Mauricio Arozi mauricioarozi at gmail.com
Mon Feb 9 06:36:40 CST 2009

Am I helpless?

On Sat, Feb 7, 2009 at 2:37 AM, Mauricio Arozi <mauricioarozi at gmail.com>wrote:

> 'I'm trying to *import/export keys* from/to Crypt_RSA<http://pear.php.net/package/Crypt_RSA>,
> using PyCrypto <http://www.dlitz.net/software/pycrypto/>. '
> and
> 'I've made some examples to find out why *it's not working*'
> I want to import/export keys directly from/to Crypt_RSA<http://pear.php.net/package/Crypt_RSA>and
> PyCrypto <http://www.dlitz.net/software/pycrypto/>." I'm not able to use
> the same keys, so I can't verify signatures, or encrypt/decrypt stuff from
> php to python and vice-versa. I want to know how to do it, actually I only
> need a way to go with php, I don't depend on Crypt_RSA, it was already there
> only.
> So in simple words, I only need to be able to encrypt/decrypt sign and
> verify signs on php and python, simultaneously, if possible, using RSA algo.
> On Fri, Feb 6, 2009 at 10:56 PM, Mads Kiilerich <mads at kiilerich.com>wrote:
>> Mauricio Arozi wrote, On 02/06/2009 07:53 PM:
>>> Hello,
>>> I'm trying to import/export keys from/to Crypt_RSA <
>>> http://pear.php.net/package/Crypt_RSA>, using PyCrypto <
>>> http://www.dlitz.net/software/pycrypto/>. My problem is that while using
>>> PyCrypto to generate both public and private keys, the e(exponent?) is
>>> always the same.
>>> According to this site: http://pajhome.org.uk/crypt/rsa/rsa.html, and
>>> yet others, the e(exponent?) is used for the public key, and d for the
>>> private key.
>> Yes, many implementations and applications of RSA uses a fixed exponent -
>> very often 65537 (F4). Usually that is a good decision and no problem -
>> perhaps except for interoperability.
>> And yes, interoperability is often very hard when implementing crypto
>> stuff. IMHO an important criteria when selecting a crypto library is having
>> examples / proof of how it interoperates with other implementations.
>> It is not clear to me exactly what you are asking for, so I can't answer
>> that directly - I hope someone else can do that.
>> /Mads
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