How does RSA work

It's pretty damn cool

Euler's theorem

\text{We have } a^{\phi(m)} = 1 \mod{m}

\text{For any } a, m \text{ such that gcd}(a, m)=1

This is fundamental for RSA. We will see that it's the reason why it works (not the reason it's secure however)

a^{\phi(m)} = 1 \mod{m}

We care because 1 means we cycle.

\text{Euler's theorem}

And let's use numbers cos why not

\text{Let } a = 2, m = 7

\text{By Euler's theorem, we have } 2^{\phi(7)} = 1 \mod{7}.

(\text{Since 7 is prime, } \phi(7) = 7 - 1 = 6)

2^6 = 1 \mod{7}

2^7 = \underbrace{2^6}_\text{1} \cdot 2 = 2^1 \mod{7}

2^{15} = \underbrace{2^{6 \cdot 2}}_\text{1} \cdot 2^3 = 2^3 \mod{7}

a^t = a^{(t \mod{\phi(m)})} \mod{m}

Gotta be inverses of each other right?

\text{dec}(\text{enc}(a)) = a

From the last slide

a^t = a^{(t \mod{\phi(m)})} \mod{m}

Recalling those basic rules

a^1 = a

(a^t)^s = a^{t \cdot s}

Can we come up with something?

\text{Because } \phi(N) \text{ is bloody hard to compute}

(incomplete)

Mathieu Paturel

30 July 2020

Finite Maths (MATH2400 <3)

By Mathieu Paturel