RSA in pure number theory
Binomial Theorem Given $a, b \in \mathbb{R}, n \in \mathbb{N}$, then $$ (a + b)^n = \sum_{k = 0}^{n} \binom{n}{k}a^k b^{n-k} $$ Proposition Let $p$ be prime. $\forall a, b \in \mathbb{Z}$, we have: $$ (a+b)^p \equiv a^p + b^p \mod p $$ Proof It suffices to show that $\binom{p}{k} \equiv 0 \mod p$ if $0 < k < p$. Observe: $p \mid p!, p \nmid k!, p \nmid (p-k)!$, hence $p \mid \binom{p}{k}$. $\blacksquare$ ...