Digital Media Concepts/RSA (cryptosystem)
What is RSAEdit
It is one of the first asymmetric cryptography algorithms which uses a pair of keys to encrypt and decrypt data. This algorithm is now widely used to transmit sensitive data over insecure network like the Internet.^{[1]}
History^{[1]}Edit
Before mid1970s, people used symmetric cryptography algorithms to encrypt data. With these algorithms, data encrypted with a particular cryptographic key could only be decrypted with the same key. Anyone without the key cannot decrypt the data. Therefore people could safely send sensitive information through insecure communication channel. However, people could not find a way to safely exchange their keys between them. Asymmetric cryptography algorithms were invented to solve this problem, and RSA is one of them. In April 1977, RSA was invented by three people from Massachusetts Institute of Technology, including two computer scientists, Ron Rivest, Adi Shamir, and a mathematician Leonard Adleman, and was later publicized in August of the same year.
RSA is now in public domain, and can be freely implemented and used by anyone.^{[2]}
Steps^{[3]}Edit
Generate the keys^{[4]}Edit
As an asymmetric cryptography algorithm, RSA cryptosystem involves two keys, the public key and the private key. Data encrypted by one key can only be decrypted with the other.
 Randomly choose two prime numbers, and .
 Multiply them together:
 Get the least common multiple of and :
 Randomly choose a positive integer which is less than and coprime to .
 Calculate modular multiplicative inverse of modulo . Which means, find a positive integer which is less than that satisfies:
 Now the key pair is generated. is the private key, and is the public key
Sample CodeEdit
Here's a sample implementation of generating RSA keys written in Python 3.6
Code


import secrets
def gcd(x: int, y: int) > tuple:
''' Extended Euclidean Algorithm
return: a, b, gcd
ax + by = gcd
'''
def f(x: int, y: int) > tuple:
l = []
while y != 0:
q, r = divmod(x, y)
l.append(q)
x, y = y, r
x, y, g = (1, 0, x) if x >= 0 else (1, 0, x)
while l:
x, y = y, x  l.pop() * y
return x, y, g
# below is a recursive approach. easier to understand
# but it may exceeds python recursion limit with large numbers
# if y == 0:
# return (1, 0, x) if x >= 0 else (1, 0, x)
# quot, rem = divmod(x, y) # rem + quot * y = x . . . . . (1)
# _a, _b, _gcd = f(y, rem) # _a * y + _b * rem = _gcd . . (2)
# return _b, _a  quot * _b, _gcd # < plug rem from (1) into (2)
assert isinstance(x, int) and isinstance(y, int), 'Integers expected'
return f(x, y)
def miller_rabin(n: int, k: int=100):
''' MillerRabin primality test
https://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test :
Input #1: n > 3, an odd integer to be tested for primality
Input #2: k, the number of rounds of testing to perform
Output: “composite” if n is found to be composite,
“probably prime” otherwise
write n as 2^r·d + 1 with d odd (by factoring out powers of 2 from n − 1)
WitnessLoop: repeat k times:
pick a random integer a in the range [2, n − 2]
x ← a^d mod n
if x = 1 or x = n − 1 then
continue WitnessLoop
repeat r − 1 times:
x ← x^2 mod n
if x = n − 1 then
continue WitnessLoop
return “composite”
return “probably prime”
'''
assert isinstance(n, int) and isinstance(k, int) and k > 0
if n < 4:
return n > 1
if n & 1 == 0: # even numbers
return False
r = 0
n_1 = n  1
d = n_1
while d & 1 == 0:
d >>= 1
r += 1
for i in range(k):
a = secrets.randbelow(n  3) + 2
x = pow(a, d, n)
if x == 1 or x == n_1:
continue
for j in range(r):
x = pow(x, 2, n)
if x == n_1:
break
else:
return False
return True
def get_prime(bits: int) > int:
''' Get a pseudoprime with given bits
This function randomly generates a number with given bits, and test its
primality using the millerrabin test.
'''
assert isinstance(bits, int) and bits > 1
low = 1 << bits  1
while True:
r = secrets.randbits(bits  1) + low
if miller_rabin(r):
return r
def generate_rsa_keys(bits: int):
''' Generate RSA key pair
It returns e, d, n, which makes
(m^e mod n)^d mod n == m
'''
def lcm(x: int, y: int) > int:
a, b, g = gcd(x, y)
return x * y // g
assert isinstance(bits, int) and bits > 5
p = q = 0
while p == q:
p = get_prime(bits // 2)
q = get_prime(bits // 2)
n = p * q
l = lcm(p  1, q  1)
_g = 0
while _g != 1:
e = secrets.randbelow(l  3) + 3 # [3, l)
d, _not_used, _g = gcd(e, l)
if d < 0: # Got negative d
d += l
return e, d, n

Send the public key to othersEdit
Let's say Alice and Bob want to have a secure communication. They may exchange their public keys without encryption. After that, a sender should always encrypt the data with the receiver's public key before sending it.
For example, Bob's data is encrypted with Bob's public key, and only people who know Alice's private key can decrypt the data. So Alice is the only person that meets this requirement. The same works with Bob.
Encryption and Decryption^{[4]}Edit
 Let's say the original data is an positive integer which should be less than . Encrypted data, positive integer , can be generated with a public key:
 The encrypted data can be decrypted with the corresponding private key:
Proof of correctness^{[4]}Edit
Proof


Because the encrypted data is calculated as:
When decrypting the encrypted data , the result is:
If equals to the original unencrypted data , then this algorithm is correct.
It says when and are coprime ( ), if both of these statements are true:
Then this is also true:
As and are both prime numbers, they are obviously coprime. So if both of these statements could be proved to be true:
Then this is also true:
Then the correctness of RSA algorithm could be proved:
To accomplish this, Fermat's little theorem is necessary. It says, if is a prime, and is not a multiple of ( ), then this statement is true:
As the relationship between and is unknown, the problem need to be divided into two cases.
Because:
As is a multiple of both and , both of these statements are true:
So can be expressed as:
Therefore:

Safety^{[5]}Edit
The public key, which is and , can be known to everyone. As far as is kept private, no one except the private key owner is able to decrypt the data encrypted by the public key. However, number has two prime factors and . If one can find and by factoring , then this person can also find out , and eventually . So the problem of how safe RSA is, is equivalent to how hard it is to factor .
It is proved that, currently with traditional (nonquantum) computer, factoring a big number is an NP problem. It may or may not be an NPcomplete problem. RSA remains safe when integer factorization can't be solved in polynomial time.
External LinksEdit
The original paper published by its inventors could be found here.
See alsoEdit
ReferencesEdit
 ↑ ^{1.0} ^{1.1} Sheposh, Richard. 2017. “RSA (Cryptosystem).” Salem Press Encyclopedia of Science.
 ↑ "RSA Security Releases RSA Encryption Algorithm into Public Domain  RSA, The Security Division of EMC". web.archive.org. 20071120. Retrieved 20190930.
 ↑ "What is RSA algorithm (RivestShamirAdleman)?  Definition from WhatIs.com". SearchSecurity. Retrieved 20190930.
 ↑ ^{4.0} ^{4.1} ^{4.2} Rivest, R. L.; Shamir, A.; Adleman, L. (19782). "A Method for Obtaining Digital Signatures and Publickey Cryptosystems". Commun. ACM 21 (2): 120–126. doi:10.1145/359340.359342. ISSN 00010782. http://doi.acm.org/10.1145/359340.359342.
 ↑ "cryptography  How safe and secure is RSA?". Stack Overflow. Retrieved 20191001.