# RSA What is interesting about an RSA key: `e` is a constant, often it is 65537 `n` is the modulus, `p * q = n` through factorization 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) $$ \phi(n) = (p-1)(q-1) $$ and further $$ 1 < \phi < n $$ Encryption, public key `e` is a prime between 2 and phi $$ 2 < e < \phi $$ Decryption, private key `d` $$ d\ e\ mod\ \phi(n) \equiv 1 $$ $$ d\ e \equiv 1\ (mod\ \phi(n)) $$ `d` is the modular inverse of e and phi and makes the private key. $$ Cipher = msg^{d}\ mod\ \phi $$ $$ Cleartext = cipher^{e}\ mod\ \phi $$ --- `e` and `d` may be found through the following Python snippets ```python possible_e = [] for i in range (2, phi): if gcd(n, i) == 1 and gcd(phi, i) == 1: possible_e.append() ``` ```python possible_d = [] for i in range (phi + 1, phi + foo): if i * e mod phi == 1 : possible_d.append() ``` ## 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 ## RSA PublicKey Extraction ### Extract n and e from RSA public key ```python from Crypto.PublicKey import RSA with open("./id_rsa.pub", 'r') as _f: pub_k = RSA.importKey(_f.read()) print(f"n:\n{pub_k.n}\n") print(f"\ne:\n{pub_k.e}\n") ``` ### Extract p and q from PublicKey Modified from [d4rkvaibhav](https://github.com/murtaza-u/zet/tree/main/20220808171808/README.md) ```python from Crypto.PublicKey import RSA with open("./id_rsa.pub", 'r') as _f: pub_k = RSA.importKey(_f.read()) def isqrt(n): x=n y=(x+n//x)//2 while(y