1
$\begingroup$

Is there a known weakness in RSA when the value of modulus (n) is very large?

n = 2466 digits
e = 65537
c = 2466 digits

n cannot be factorized. I checked on factordb.com and no factors were present.

Both, modulus and ciphertext have the same number of digits. Is this not considered secure?

What kind of attacks can be used against such a crypto system?

$\endgroup$
  • $\begingroup$ Is this related to some CTF competition or something? I'm just wondering because just a couple of hours earlier we got another remarkably similar question that has since been deleted. $\endgroup$ – Ilmari Karonen Jun 5 at 11:45
1
$\begingroup$

2466 digits for an RSA modulus is somewhat large, but not unseen. A larger RSA modulus, when it is properly sampled, can only lead to a better security guarantee. The ciphertext in RSA is $c = M^e \bmod n$, where $M$ is something derived from the input message $m$ - hence, it is perfectly normakl for $c$ to have the same bit length as $n$ (anything significantly smaller would be a weakness, anything significantly larger... would not be a valid ciphertext). This choice of $e$ is also the most common one. Hence, there is nothing wrong with your proposed choice of parameters, provided that $n$ was indeed generated as the product of two random length-$1233$ strong primes.

n cannot be factorized. I checked on factordb.com and no factors were present.

Note, however, that this gives you essentially zero indication about the actual security of this modulus. It is only a secure modulus if it was generated as a product of two equal-length random strong primes, with a good random number generator. $n$ could be a very weak RSA modulus if it was not generated this way, yet you would not immediately find the factors with an online tool.

$\endgroup$
  • $\begingroup$ I believe that when he said both were 2466 digits long, he meant decimal digits; this translates to circa 8192 bits... $\endgroup$ – poncho Jun 5 at 11:33
  • $\begingroup$ woops, misread digits for bits. Will correct that then. $\endgroup$ – Geoffroy Couteau Jun 5 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.