2
$\begingroup$

On some lecture slides regarding to RSA-Encryption, the formula for calculation of the private key is given as $d = e^{-1} \equiv e^{\phi(N)-1} \mod \phi(N)$. The second equation is justified by the fact that $gcd(e,\phi(N))=1$.

My questions:

1.How does the second equality come about? I know, that Eulers Theorem shows, that $a^{\phi(b)} \equiv 1 \mod b$ for coprime $a,b$. So why we dont need to write $e^{\phi(\phi(N))-1} \mod \phi(N)$?

  1. With this forumula we can calculate $d$ easily, aren't we? So it is common, to use the Euklid's-Algorithm? It needs much more calculation steps?

I hope someone can help me to understand this better.

$\endgroup$
0

1 Answer 1

2
$\begingroup$

1.How does the second equality come about? I know, that Eulers Theorem shows, that $a^{\phi(b)} \equiv 1 \mod b$ for coprime $a,b$. So why we dont need to write $e^{\phi(\phi(N))-1} \mod \phi(N)$?

Actually, you are correct - the slides were wrong.

  1. With this forumula we can calculate $d$ easily, aren't we? So it is common, to use the Euklid's-Algorithm? It needs much more calculation steps?

Well, to use the relation you cite, we would need to know the value $\phi(\phi(N))$, which means we would know the complete factorization of $p-1, q-1$. Depending on the prime generation method, we might not (and the ones that would give us that are take much more time or are more complex). Euclid's algorithm, on the other hand, doesn't need that information, and so it's much simpler.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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