# RSA

* `p * q = n`
* Coprime Phi is calculated either by [Euler Totient](https://en.wikipedia.org/wiki/Euler's_totient_function) or [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) via [euclidean algorithm](https://crypto.stanford.edu/pbc/notes/numbertheory/euclid.html) 

$$
1 < $\phi$ < n
$$

* There is also 
$$
$\phi$ = (p-1) * (q-1)
$$$

* Encryption, public key `e` is a prime between 2 and phi  
$$
2 < e < $\phi$
$$

```python
possible_e = []
for i in range (2, phi):
    if gcd(n, i) == 1 and gcd(phi, i) == 1:
        possible_e.append() 
```

* Decryption, private key `d` 
$$
d * e mod $\phi$ = 1
$$

```python
possible_d = []
for i in range (phi + 1, phi + foo):
    if i * e mod phi == 1 :
       possible_d.append()
```
* \\( Cipher = msg ** d mod $\phi$ \\)
* \\( Cleartext = cipher ** e mod $\phi$ ) 

## Euklid

Just a short excourse:  
A greatest common divisior out of an example a = 32 and b = 14 would be the groups of the following divisors
```sh
a = 32, b = 24
a = {1, 2, 4, 8, 16}
b = {1, 2, 3, 8, 12}
gcd(a,b) = 8
```

### Greatest Common Divisor (GCD)

Two values are prime and have themselves and only `1` as a divisor are called coprime.
To check if a and b have a greatest common divisor do the euclidean algorithm.
```python
def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)
```

### Extended GCD

#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.

So a porperty of quad res are, if Quadratic Residue $QR = 1$ and Quadratic NonResidue $QN = -1$

$$
QR * QR = QR\\
QR * QN = QN\\
QN * QN = QR\\
$$

## Legendre

$$
\frac{a}{p} =
\begin{cases}
1, & if\ a\ quadratic\ residue\ mod\ p\ and\ not\ a\ \equiv\ 0\ (mod\ p),\\
-1, & if\ a\ is\ a\ non\ residue\ mod\ p,\\
0, & if\ a\ \equiv 0\ (mod\ p)\\
\end{cases}
$$

$$
\frac{a}{p} \equiv a^{p-1/2}\ (mod\ p)\ and\ \frac{a}{p} \in \{-1,0,1\}
$$

* Legendre Symbol test via Python with 
```python
pow(a,(p-1)/2,p)
```

[Finding the square root of integer a which is quadratic residue](http://mathcenter.oxford.emory.edu/site/math125/findingSquareRoots/)

* Given $p \equiv 3\ mod\ 4$ the square root is calculated through
```python
pow(a,((p+1)//4),p)
```

## Tonelli-Shanks - Modular Square Root

* Find elliptic curve co-ordinates
* Precondition: modulus is not a prime
* TBD

## Links

* [Encryption+Decryption](https://www.cs.drexel.edu/~jpopyack/Courses/CSP/Fa17/notes/10.1_Cryptography/RSA_Express_EncryptDecrypt_v2.html)
* [Extended GCD](http://www-math.ucdenver.edu/~wcherowi/courses/m5410/exeucalg.html)