[Crypto] Is the limited set of exponents (e) in RSA a backdoor?
Why only {3, 5, 17, 257, 65537}?
Some says it's for performance reasons(Fermat primes).
Is it a possibility that some sort of "rainbow tables" have been computed for these values ?
1 comments
They are primes of low hamming weight, so the exponentiationd you have to do are cheaper (imagine a multiplication, long hand, where almost all of the digits are 0: that really helps). You actually don't want a small exponent, as it will fail to saturate the modulus (so please, don't use 3, that is a well-known bad thing to do with RSA). As for a "rainbow table", that is a concept that only applies to hashes; if you just mean a full lookup table, the time complexity of attacking RSA directly comes from the modulus, not the exponent.
Thank you for your explanation.