From d9faa96cd028d53a83fd9180e288f8294fe457f5 Mon Sep 17 00:00:00 2001
From: whx <stefan@stefan.works>
Date: Thu, 17 Nov 2022 00:12:56 +0100
Subject: [PATCH] updated rsa

---
 Cryptography/RSA.md | 32 ++++++++++++++++++++++++++++++++
 1 file changed, 32 insertions(+)

diff --git a/Cryptography/RSA.md b/Cryptography/RSA.md
index bf68040..f7dc103 100644
--- a/Cryptography/RSA.md
+++ b/Cryptography/RSA.md
@@ -49,6 +49,38 @@ def gcd(a, b):
 
 #TODO
 
+## Fermat's Little Theorem
+
+If modulus $p$ is a prime and and modulus $n$ is not a prime, p defines a finite field (ring).
+$$
+n \in F_{p} \{0,1,...,p-1\}
+$$
+
+The field consists of elements $n$ which have an inverse $m$ resulting in $n + m = 0$ and $n * m = 1$.
+
+So , $n^p - n$ is a multiple of p then $n^p \equiv n\ mod\ p$ and therefore $ n = n^p\ mod\ p$. An example
+$$
+4 = 4^{31}\ mod\ 31
+$$
+
+Further, $p$ while still a prime results in $1 = n^{p-1} mod\ p$. An example
+$$
+1 = 5^{11-1}\ mod\ 11
+$$
+
+### Modular Inverse
+
+Coming back to the modular inverse $n$, it can be found in the following way
+$n^{p-1} \equiv 1\ mod\ p$  
+$n^{p-1} * n^{-1} \equiv n^{-1}\ mod\ p$  
+$n^{p-2} * n * n^-1 \equiv n^{-1}\ mod\ p$  
+$n^{p-2} * 1 \equiv n^{-1}\ mod\ p$  
+$n^{p-2} \equiv n^{-1}\ mod\ p$
+
+## Quadratic Residue
+
+$m$ is a quadratic residue when $\pm n^2 = m\ mod\ p$ with two solutions.
+Otherwise it is a quadratic non residue.
 
 ## Links