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**RSA** is a cryptosystem, which is known as one of the first practicable **public-key cryptosystems** and is widely used for **secure data transmission**. In such a cryptosystem, the **encryption key is public and differs** from the **decryption key** which is kept secret.

In RSA, this asymmetry is based on the practical difficulty of **factoring the product of two large prime numbers**, the factoring problem. RSA stands for Ron Rivest, Adi Shamir and Leonard Adleman, who first publicly described the algorithm in 1977. Clifford Cocks, an English mathematician, had developed an equivalent system in 1973, but it wasn't declassified until 1997.

A user of RSA creates and then publishes the product of two large prime numbers, along with an auxiliary value, as their public key. The prime factors must be kept secret. Anyone can use the public key to encrypt a message, but with currently published methods, if the public key is large enough, only someone with knowledge of the prime factors can feasibly decode the message.Whether breaking RSA encryption is as hard as factoring is an open question known as the RSA problem.

RSA Algorithm Example

- Choose p = 3 and q = 11
- Compute n = p * q = 3 * 11 = 33
- Compute φ(n) = (p - 1) * (q - 1) = 2 * 10 = 20
- Choose e such that 1 < e < φ(n) and e and n are coprime. Let e = 7
- Compute a value for d such that (d * e) % φ(n) = 1. One solution is d = 3 [(3 * 7) % 20 = 1]
- Public key is (e, n) => (7, 33)
- Private key is (d, n) => (3, 33)
- The encryption of m = 2 is c = 27 % 33 = 29
- The decryption of c = 29 is m = 293 % 33 = 2

The **RSA algorithm** involves three steps:

**1. Key Generation**

**2. Encryption**

**3. Decryption**

1. The KEY GENERATIONRSA involves a

public keyanda private key. Thepublic keycan be known by everyone and is used forencrypting messages. Messages encrypted with the public keycan only be decryptedin a reasonable amount of time using theprivate key. The keys for the RSA algorithm are generated the following way:Choose two distinct

PRIME NUMBERSp and q.For security purposes, the integers p and q should be chosen at random, and should be of similar bit-length. Prime integers can be efficiently found using a primality test.

Compute n = p*q.n is used as the

modulusfor both thepublic and private keys. Its length, usually expressed in bits, is thekey length.Compute φ(n) = φ(p)*φ(q) = (p − 1)*(q − 1), where φ is

Euler'stotient function.Choose an integer

esuch that1 < e < φ(n)andgcd(e, φ(n)) = 1i.e. e and φ(n) are coprime.e is released as the public key exponent.

e having a short bit-length and small Hamming weight results in more efficient encryption – most commonly 216 + 1 = 65,537. However, much smaller values of e (such as 3) have been shown to be less secure in some settings.Determine d as d−1 ≡ e (mod φ(n)), i.e., d is the multiplicative inverse of e (modulo φ(n)).

This is more clearly stated as: solve for d given d⋅e ≡ 1 (mod φ(n))

This is often computed using the extended Euclidean algorithm. d is kept as theprivate key exponent.The

public key consistsof themodulus nand thepublic (or encryption) exponent e.The

private key consistsof themodulus nand theprivate (or decryption) exponent d, which must be kept secret.

p, q, and φ(n) must also be kept secretbecause they can be used to calculate d.

2. ENCRYPTIONAlice transmits her

public key (n, e)to Bob and keeps theprivate key secret. Bob then wishes to sendmessage Mto Alice.He first turns M into an integer m, such that 0 ≤ m < n by using an agreed-upon reversible protocol known as a padding scheme. He then computes the ciphertext c corresponding to

c = m^e * mod{n}This can be done quickly using the method of exponentiation by squaring. Bob then transmits c to Alice.

Note that at least nine values of m will yield a ciphertext c equal to m but this is very unlikely to occur in practice.

3.DECRYPTIONAlice can recover m from c by using her private key exponent d via computing

m = c^d mod{n}Given m, she can recover the original message M by reversing the padding scheme.

A WORKING EXAMPLEHere is an example of

RSA encryption and decryption. The parameters used here are artificially small, but one can also useOpenSSLto generate and examine a real keypair.Choose two distinct prime numbers, such as

p = 61 and q = 53.

Compute n = pq giving

n = 61 * 53 = 3233.

Compute the totient of the product as φ(n) = (p − 1)*(q − 1) giving

φ(3233) = (61 - 1)*(53 - 1) = 3120.

Choose any number 1 < e < 3120 that is coprime to 3120. Choosing a prime number for e leaves us only to check that e is not a divisor of 3120. Let e = 17. Compute d, the modular multiplicative inverse of e (mod φ(n)) yielding d = 2753. The

public keyis (n = 3233, e = 17). For a padded plaintext message m, the encryption function isc(m) = m^{17} * mod 3233.

The

private keyis (n = 3233, d = 2753). For an encrypted ciphertext c, the decryption function ism(c) = c^{2753} * mod 3233.

For instance, in order to encrypt m = 65, we calculate

c = 65^{17} * mod 3233 = 2790

To decrypt c = 2790, we calculate

m = 2790^{2753} * mod 3233 = 65.

Both of these calculations can be computed efficiently using the square-and-multiply algorithm for modular exponentiation.

In real-life situations the primes selected would be much larger; in our example it would be trivial to factor n, 3233 (obtained from the freely available public key) back to the primes p and q. Given e, also from the public key, we could then compute d and so acquire the private key.