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reconstruct a private key from a public key

This commit is contained in:
gurkenhabicht 2024-02-18 21:09:29 +01:00
parent b6788a4bb4
commit 513a004f9f
2 changed files with 158 additions and 26 deletions
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36
Cryptography/OpenSSL-Cheatsheet.md
View File

@ -1,24 +1,26 @@
# OpenSSL Cheatsheet
## Read X.509 Certificate
* A certificate can be read via
A certificate can be read via
```sh
openssl x509 -in $CERT -text
```
## Generate CSR
* A Certificate Signing Request needs a private alongside the request for a cert.
A Certificate Signing Request needs a private alongside the request for a cert.
This is done in the following way
```sh
openssl req -new -nodes -newkey rsa:4096 -keyout $PRIVATE_KEY -out $CERT_CSR
```
## Create an X.509 Certificate
* Create a X.509 certificate via
Create a X.509 certificate via
```sh
openssl x509 -newkey -nodes rsa:4096 -keyout $PRIVATE_KEY -out $CERT -sha256 -days 365
openssl req -new -x509 -keyout cert.pem -out cert.pem -days 365 -nodes
@ -26,7 +28,8 @@ openssl req -new -x509 -keyout cert.pem -out cert.pem -days 365 -nodes
## Extract Keys from PFX Cert
* Key and cert form PFX
Key and cert form PFX
```sh
openssl pkcs12 -in cert.pfx -nocerts -out key.pem -nodes
openssl pkcs12 -in cert.pfx -out cert.pem -clcerts -nokeys
@ -34,7 +37,8 @@ openssl pkcs12 -in cert.pfx -out cert.pem -clcerts -nokeys
## Extract & Repack PFX Cert
* Extract & Repack with another password, e.g. from `mimikatz` to `cqure`
Extract & Repack with another password, e.g. from `mimikatz` to `cqure`
```sh
openssl pkcs12 -in *.pfx -out temp.pem -nodes
openssl pkcs12 -export -out *.pfx -in temp.pem
@ -44,26 +48,32 @@ openssl pkcs12 -export -out *.pfx -in temp.pem
### Read Parameters of a RSA Key
* Show parameters of the private key
Show parameters of the private key
```sh
openssl rsa -in $PRIVATE_KEY -text -noout
```
### Create RSA Key
* Generate an OpenSSL RSA key via
Generate an OpenSSL RSA key via
```sh
openssl genrsa -out $PRIVATE_KEY 4096
```
* Generate an OpenSSl RSA public key from a private key
Generate an OpenSSl RSA public key from a private key
```sh
openssl rsa -in $PRIVATE_KEY -pubout -out public-key.pem
```
### Encrypt RSA
* Encrypt RSA current and deprecated
Encrypt RSA current and deprecated
```sh
openssl pkeyutl -encrypt -in $CLEAR_TEXT -out $CLEAR_TEXT -pubin -inkey $PUBLIC_KEY
openssl rsautl -encrypt -in $CLEAR_TEXT -out $ENCRYPTED -pubin -inkey $PUBLIC_KEY
@ -71,12 +81,14 @@ openssl rsautl -encrypt -in $CLEAR_TEXT -out $ENCRYPTED -pubin -inkey $PUBLIC_KE
### Decrypt RSA
* Decrypt a RSA cipher with the private key
Decrypt a RSA cipher with the private key
```sh
openssl pkeyutl -decrypt -in $CIPHER -out $PLAIN_TEXT -inkey $PRIVATE_KEY
```
* Deprecated version of RSA decryption is the following
Deprecated version of RSA decryption is the following
```sh
openssl rsautl -decrypt -in $CIPHER -out $PLAIN_TEXT -inkey $PRIVATE_KEY
```

144
Cryptography/RSA.md
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@ -7,12 +7,12 @@ $$
1 < \phi < n
$$
* There is also
There is also
$$
\phi = (p-1) * (q-1)
$$$
* Encryption, public key `e` is a prime between 2 and phi
Encryption, public key `e` is a prime between 2 and phi
$$
2 < e < \phi
$$
@ -24,7 +24,7 @@ for i in range (2, phi):
possible_e.append()
```
* Decryption, private key `d`
Decryption, private key `d`
$$
d * e mod \phi = 1
$$
@ -35,13 +35,16 @@ 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
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}
@ -53,6 +56,7 @@ gcd(a,b) = 8
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:
@ -62,18 +66,21 @@ def gcd(a, b):
### Extended GCD
#TODO
\#TODO
## Fermat's Little Theorem
## Fermat`s Little Theorem
If modulus $p$ is a prime and and modulus $n$ is not a prime, p defines a finite field (ring).
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 $n * m = 1$.
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
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
$$
4 = 4^{31}\ mod\ 31
$$
@ -97,7 +104,8 @@ $n^{p-2} \equiv n^{-1}\ mod\ p$
$m$ 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 $QN = -1$
So a porperty of quad res are, if Quadratic Residue $QR = 1$ and Quadratic
NonResidue $QN = -1$
$$
QR * QR = QR\\
@ -120,14 +128,16 @@ $$
\frac{a}{p} \equiv a^{p-1/2}\ (mod\ p)\ and\ \frac{a}{p} \in \{-1,0,1\}
$$
* Legendre Symbol test via Python with
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
Given $p \equiv 3\ mod\ 4$ the square root is calculated through
```python
pow(a,((p+1)//4),p)
```
@ -138,6 +148,116 @@ pow(a,((p+1)//4),p)
* 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<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
```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<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'))
```
## Links
* [Encryption+Decryption](https://www.cs.drexel.edu/~jpopyack/Courses/CSP/Fa17/notes/10.1_Cryptography/RSA_Express_EncryptDecrypt_v2.html)