1. The following problem explains RSA encryption, which is currently used to encrypt information sent
over the internet, like credit card numbers. (Although elliptic key cryptography has been taking over
recently.) The ingredients are as follows:
• Two distinct three-hundred-digit primes $p$ and $q$. Let $n = pq$.
• An encryption key $e$ so that $gcd(e, (p-1)(q - 1)) = 1$.
• A secret message $m$ that Alice wants to send to Bob.
Since the evil Connor is watching their communication channel, Alice has to encrypt her message first. First
Alice sends Bob $n$ and the encryption key $e$. Then, she sends him the coded message:
$c \equiv m^e \pmod{n}$.
Bob computes the decryption key $d$ as follows:
$d \equiv e^{-1} \pmod{(p-1)(q - 1)}$
which exists because $gcd(e, (p-1)(q-1)) = 1$. The goal is to show that $c^d \pmod{n}$ computes the original
message $m$.
(a) Show that $de = k(p-1)(q-1) + 1$ for some integer $k$.
(b) Use Fermat's Little Theorem to show that $c^d \equiv m \pmod{p}$. You may assume that $gcd(m,p) = 1$. Since
$p$ is a large prime, $gcd(m, p) \ne 1$ is unlikely in the extreme.
(c) Show that $c^d \equiv m \pmod{q}$. You may again assume that $gcd(m, q) = 1$.
(d) Use the Chinese Remainder Theorem to show that $c^d \equiv m \pmod{n}$. Therefore, Bob can uncover the
original message using the decryption key.
(e) Explain why Connor, who only sees $n$, $e$, and $c$, will have a tough time finding $m$. (He doesn't know $p$
and $q$, which are three-hundred-digit primes.)
