# Tag Info

## Hot answers tagged rsa

16

Suppose an RSA public key is still $(e, N)$ and the private key is still $d$. We have $Enc = (m*e) \text{ mod } N$ and $Dec = (c*d) \text{ mod } N$. The correctness of the scheme depends on the fact that $Dec(Enc(m)) = (m*e*d) \text{ mod } N = m \text{ mod } N$. This implies $e*d = 1 \text{ mod } N$. It is thus trivial to compute the private key given only ...

7

Assumption: $e$ and $d$ are prime with $N$. If we use the multiplication instead of the power. We have $d \equiv e^{-1} \pmod N$ : \begin{align}c \times d &= m &\mod N\\ m \times e \times d &= m &\mod N\\ e \times d &= 1 &\mod N \end{align} (It has to work for any $m$ so we can suppose $m$ prime). $e$ and $N$ are publicly known, to ...

6

No, it is not easy! RSA is based on the difficulty of factoring the product $n=pq$ of two large prime numbers. But if you know $\varphi(n)$ for plain RSA you can compute the secret exponent $d=e^{-1}\bmod \varphi(n);\;$ and you can factor $n$ from the two equations $n=pq,\;\varphi(n)=(p-1)(q-1)$.

3

For private key operations you need at least $n$ (the modulus) and $d$ (the private exponent). The primes $p$ and $q$ let you calculate those – or use some shortcuts for quicker computation – so they also suffice. In practice RSA keys often include all of those values, to avoid having to compute them as needed and to allow for optimized and unoptimized ...

2

I will assume that Alice uses $D' = E^{-1}(\varphi(N) - 1)$ with $E^{-1}$ the inverse of $E$ modulo $\lambda(N)$. The question uses a very specific definition of $D'$, which requires the assumption that $E$ divides $\varphi(N) - 1$. But since such an $E$ is necessarily coprime to $\varphi(N)$, it is a special case of the more general definition of $D'$ ...

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