Why is $P^{ed} \bmod N = P^{1} \bmod N$ if $e$ and $d$ are modular inverses of $\phi(N)$.

Asked May 10 '21 at 18:44

Active May 10 '21 at 18:58

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Let $d$ be the inverse of $e \bmod \phi(N)$.

Therefore,

$de \equiv 1 \pmod{\phi(N)} $

$d \equiv e^{-1} \pmod{\phi(N)} $

In RSA encryption, we can decrypt messages because $P^{ed} \bmod N = P^{1} \bmod N$

I don't understand why $P^{ed} \bmod N = P^{1} \bmod N$ if $e$ and $d$ are modular inverses of $\phi(N)$. What property is this?

edited May 10 '21 at 18:58

Arturo Magidin

asked May 10 '21 at 18:44

2

Because $a^{\phi(N)}=1$ modulo $N$ by Euler, see here, for example. – Dietrich Burde May 10 '21 at 18:48
I still don't understand because $ed \neq \phi(N)$ – inquisitivemongoose May 10 '21 at 19:13
Note 1 raised to any power is 1 . – Roddy MacPhee May 10 '21 at 19:34

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