📐 RSA Correctness

For properly chosen $e$ and $d$, decryption correctly recovers the original message: $m^{ed} \\equiv m \\pmod{n}$.

Proof: **Setup:** Let $n = pq$ for distinct primes $p, q$. Let $e$ be coprime to $\\varphi(n) = (p-1)(q-1)$, and let $d = e^{-1} \\pmod{\\varphi(n)}$.

This means $ed = 1 + k \\cdot \\varphi(n)$ for some integer $k$.

**Proof:**

We need to show $m^{ed} \\equiv m \\pmod{n}$ for any message $m$ with $0 \...

From: Cryptography Math

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