5.6 KiB

Raw Blame History

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 or greatest common divisor via euclidean algorithm


\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

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

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

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.

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

pow(a,(p-1)/2,p)

Finding the square root of integer a which is quadratic residue

Given p \equiv 3\ mod\ 4 the square root is calculated through

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

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

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<x):
        x=y
        y=(x+n//x)//2
    return x

def fermat(n):
    t0=isqrt(n)+1
    counter=0
    t=t0+counter
    temp=isqrt((t*t)-n)
    while((temp*temp)!=((t*t)-n)):
        counter+=1
        t=t0+counter
        temp=isqrt((t*t)-n)
    s=temp
    p=t+s
    q=t-s
    return p,q

p,q = fermat(pub_k.n)
print(f"\np: {p}\n")
print(f"\nq: {q}\n")
print(f"\np-q: {p-q}\n")

Generate PrivateKey

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<x):
        x=y
        y=(x+n//x)//2
    return x

def fermat(n):
    t0=isqrt(n)+1
    counter=0
    t=t0+counter
    temp=isqrt((t*t)-n)
    while((temp*temp)!=((t*t)-n)):
        counter+=1
        t=t0+counter
        temp=isqrt((t*t)-n)
    s=temp
    p=t+s
    q=t-s
    return p,q

def extended_euclid(a, b):
    if a == 0:
        return b, 0, 1
    else:
        g, y, x = extended_euclid(b % a, a)
        return g, x - (b // a) * y, y

def modular_inverse(e, phi):
    g, x, y = extended_euclid(e, phi)

    if g != 1 :
        raise Exception("No modular inverse")
    else:
        return x % phi

p,q = fermat(pub_k.n)
phi = (p-1) * (q-1)
d = modular_inverse(pub_k.e, phi)


print(f"\np: {p}\n")
print(f"\nq: {q}\n")
print(f"\np-q: {p-q}\n")
print(f"\nd: {d}\n")

priv_k = RSA.construct((pub_k.n, pub_k.e, d))
with open ("./priv_id_rsa", "wb") as _f:
    _f.write(priv_k.export_key('PEM'))

5.6 KiB Raw Blame History

RSA