Tell me more ×
Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. It's 100% free, no registration required.
up vote 0 down vote favorite

Comparing to this question, assume $C, M \in \mathbb Z^*_{n^2}$, $e \ge 3$, is it hard to compute $M$ that satisfies $C=M^e \mod n^2$ when $C$ and $(n, e)$ are given?

share|improve this question
3  
Hint: $C=M^e \bmod n^2\implies C\equiv M^e \pmod n$. – fgrieu Feb 25 at 7:19
@fgrieu Thank you for the hint! – phan Feb 26 at 4:09

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.