# Re: Generating RSA key with different public exponent

*From*: "Valery Pryamikov" <valery@xxxxxxxxx>*Date*: 6 Sep 2006 12:08:41 -0700

Hi,

unfortunately, in your case you'll need to generate keypair yourself.

BTW: 3 is a tricky public exponent. If you use 3 as public exponent you

have to be sure that you use OAEP for encryption and PSS for signature.

Also you have to be absolutely sure that you or your partners don't

have implementation errors. Daniel Bleichenbacher is continiously

demonstrating attacks on buggy implementations with exponent 3 on PKCS

1.5 padding since 1998. This year, on Crypto 2006 he has demonstrated

how to manually forge sighature with exponent 3 and a buggy verifier,

that he has actually found at least one open source implementation

having a bug allowing the attack. Daniel is trying to convince people

to stop using RSA keys with exponents of 3. Even if your own

implementation is not vulnerable to this attack, there's no telling

what the other guy's code may do. And he is the one relying on your

signature...

-Valery.

http://www.harper.no/valery

Kris wrote:

I need to generate RSA keys using a different public exponent from the

default one used by the RSACryptoServiceProvider (65537, {1,0,1}). The

reason I need to do this is that I am currently writing software which

communicates with a device that annoyingly assumes the public exponent

of the 1024 bit modulus you send it is 3 (non-configurable on the

device).

Does anyone know if this is possible using the .NET crypto classes or

am I doomed to write a lot of code to generate keys myself?

Many thanks

Kris Sheglova

.

**Follow-Ups**:**Re: Generating RSA key with different public exponent***From:*Eugene Mayevski

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