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The RSA Algorithm

Rocky K. C. Chang, March 2014

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Outline Trapdoor one-way function The RSA algorithm Some practical considerations RSA’s security Some pitfalls of RSA

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Trapdoor one-way function Suppose n = pq, where p and q are large

primes. Consider f(m) = me mod n. For certain values of e and that n is large

enough, f(m) is a one-way function. It is computationally infeasible to obtain m based

on the knowledge of n, e, and f(m). However, with the knowledge of a certain

trapdoor, the inversion is easy to do. The trapdoor for RSA is the factorization of n (i.e.,

the knowledge of p and q).

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The RSA algorithm Let n = pq, where p and q are primes. Note

that n is a composite number. Let M = C = Zn = {0, 1, 2, …, n–1}. K = {(n, p, q, d, e): e d 1 (mod (n))}.

We will see that (n) = (p–1)(q–1). For K = (n, p, q, d, e), define

EK(m) = me mod n, and DK(c) = cd mod n, where m, c Zn.

The (n, e) comprise the “public key.” The (p, q, (n), d) comprise the “private key.”

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To probe further Both encryption and decryption involve

modulo multiplications. Since n is a composite, Zn is not a group under

modulo multiplication, i.e., the inverse may not exist. Z*

n = {a Zn: gcd(a,n) = 1}. Zn \ Z*

n = {a Zn: gcd(a,n) > 1}. How many elements in Z*

n? We denote the number of elements by (n). Recall that (n) is used in determining d and e.

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The value of (n) Note that gcd(a,n) = 1 iff gcd(a,p) = 1 and

gcd(a,q) = 1. There are q numbers in Zn that satisfy a mod p =

0: {0, p, 2p, …, (q–1)p}. There are p numbers in Zn that satisfy a mod q =

0: {0, q, 2q, …, (p–1)q}. Therefore, the total number of numbers in Zn that

their gcd(a,n) > 1 is p+q–1. Thus, (n) = pq – (p+q–1) = (p–1)(q–1). Use the well-known result (in slide 28 of the

prelude slides) that if b Z*n, then b(n) 1 (mod

n). Therefore, a(p–1)(q–1) 1 (mod n), for a Z*

n.

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For example, Let p = 3, q = 5. Therefore, n = 15 and (p–1)(q–1)

= 8. For any a {0, 3, 5, 6, 9, 10, 12}, a8 ! 1 (mod

15). For any a {1, 2, 4, 7, 8, 11, 13, 14}, a8 1 (mod

15), e.g., 24 1 (mod 15). 42 1 (mod 15). 74 1 (mod 15). …

Note that primitive elements may not exist in Z*n,

because n is not a prime.

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The relationship between e and d The values of e and d have to satisfy

e d 1 (mod (p–1)(q–1)). Recall that d exists iff gcd(e,(p–1)(q–1)) = 1 (slide 17 of

the prelude slides). For example, p = 101 and q = 113.

n = pq = 11413. (n) = (p–1)(q–1) = 11200 = 26527. Pick e = 3533, which is not divisible by 2, 5, or 7. Use the extended Euclidean algorithm to compute d = e-1

mod 11200 = 6597. To encrypt m = 9726, compute 92763533 mod 11413 =

5761. To decrypt c = 5761, compute 57616597 mod 11413 =

9726.

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DK(EK(m)) = m? Recall that ed 1 (mod (n)). In other words, ed = t(n)+1, where t is a

nonnegative integer. Part 1: Let’s consider an m Z*

n. (me)d mt(n)+1(mod n). (me)d (m(n))tm (mod n). (me)d (1)tm (mod n). (me)d m (mod n).

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DK(EK(m)) = m? Part II: Let’s consider an m Zn \ Z*

n. Using the Chinese Remainder Theorem, m mod n can be

uniquely represented by (m mod p, m mod q). Note that either the following is true:

m mod p = 0 and m mod q = 0 or m mod p = 0 and m mod q 0 or m mod p 0 and m mod q = 0.

For m mod p = 0 and m mod q = 0, med mod p = 0 and med mod q = 0. Therefore, med m (mod p) = 0 and med m (mod q) = 0.

For those cases where m mod p = 0 or m mod q = 0, Say m mod p = 0 or m mod q 0, By the CRT, med mod n can be represented by (0, med mod q). Using the previous two results, (0, med mod q) is equal to (0, m mod q).

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Digital signing using RSA To sign a message m, Alice computes s = md

mod n. The pair (m,s) is a signed message. To verify the signature, anyone who knows the

public key can verify that se m mod n, the message itself.

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Some practical considerations

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Generating the RSA parameters1. Generate 2 large primes, p and q (each with

size k/2 bits).2. n (k ≥ 2048 bits) pq and (n) (p–1)(q–

1).3. Choose a random e (1 < e < (n)) such that

gcd(e,(n)) = 1.4. d = e-1 mod (n).5. Publish (n,e) and safeguard the secret (p, q,

(n), d).

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Generating the RSA parameters1. Need an efficient algorithm to generate a

large prime. Rabin-Miller test determines whether an odd

integer n is prime.2. Find 2 large primes.3. Use the Euclidean algorithm to make sure

that gcd(e,(n)) = 1.4. Use the extended Euclidean algorithm to

compute d = e-1 mod (n).

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Practical considerations Usually fix the value of e, e.g., e = 3 for

signatures and e = 5 for encryption. There are pitfalls when one is using the same

exponent for both encryption and signatures. Therefore, p – 1 and q – 1 cannot be multiples of 3

or 5. Smaller exponent for signatures (why?) Some problems with small exponents (to be

discussed shortly). Other common values for e are 17 and 65537.

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RSA’s (in)security

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The RSA’s security An obvious attack against RSA is to factor n.

If this can done, then obtain p and q. Compute (p–1)(q–1). Compute e-1 mod (p–1)(q–1) = d.

Roughly speaking, breaking the RSA algorithm is as difficult as factoring n. The “current” factoring algorithms are able to

factor numbers having up to 512 bits. On the safe side, n ≥ 2048 bits to make the

factoring problem computationally infeasible to solve.

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The RSA’s security Moreover, if one can obtain (n), one can obtain

other elements in the private key. First of all, one can obtain p and q by solving

n = pq and (n) = (p–1)(q–1).

The solution for p is given by p2 – (n – (n) + 1)p + n = 0.

In other words, if one can compute (n), one can factor p and q.

Lastly, what happen if one can obtain the value of d? n can be factored in polynomial time using a randomized

algorithm.

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Pitfalls using RSA Problem 1: If Alice signs 2 messages m1 and

m2. Eve can compute Alice’s signature on m3 = m1m2 mod n. Original signatures: m1

d and m2d.

Eve can produce the signature for m3 by multiplying m1

d and m2d.

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Pitfalls using RSA Problem 2: When RSA is used to encrypt a very

small message m. E.g., if e = 5 and m < n1/5, then me = m5 < n. Therefore,

no mod n operation needed. Simply take a fifth root of c to recover m! For example, if encrypting a 256-bit key using RSA, the

encrypted key is less than 22565 = 21280 << 22048 if n is a 2048-bit integer.

The main problem is the existence of a structure in the numbers that RSA operates on.

A possible approach is to use an encoding function to destroy the structure as much as possible.

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Message encryption using RSA Using RSA to encrypt a message is almost

never practiced. The size of the message is limited by the size of n.

Instead, choose a random secret key K, and encrypt K with the RSA key. The message encryption is based on secret key

cryptosystem, Sending Ke mod n, EK(m).

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Message encryption using RSA A better approach is:

Choose a suitable random number r {0, 1, …, n–1}.

Set K = h(r), where h() is some hash function. Send re mod n and EK(m).

Advantages: There is no structure in r. The hash function ensures that no structure

between r’s propagates to structure in the K’s.

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Digital signatures using RSA Problem: remove the structures of the

messages that are signed. Use a hash function to hash the messages.

The hash function’s output (e.g., 256 bits) is small compared with the size of n (e.g., 2048 bits). Cannot use the hash function output directly in

RSA.

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Digital signatures using RSA A solution is to use a pseudorandom mapping

to expand h(m) to a random number s {0, 1, …, n – 1}.

If you ask Alice to sign a number of messages m1, m2, …, mi. Eve can get hold of the (m, s), but the values of s

are effectively random. Thus, the information does not help forge Alice’s

signature.

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The RSA Lab’s public-key cryptography standard PKCS #1 for RSA or RFC 3447 covers

Data conversion primitives: a text <-> a non-negative integer

Cryptographic primitives Encryption schemes

RSAES-OAEP (for new applications) – cryptographic primitives + Bellare and Rogaway's Optimal Asymmetric Encryption scheme

RSAES-PKCS1-v1_5 (for existing applications) – cryptographic primitives + a PKCS1-v1_5 encoding method

Digital Signature schemes RSASSA-PSS (for new applications) – cryptographic primitives

+ a probabilistic signature scheme-based encoding method RSASSA-PKCS1-v1_5 (for existing applications) – cryptographic

primitives + a PKCS1-v1_5 encoding method

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Conclusions RSA can be used for encryption as well as digital

signatures. The security of RSA lies on the difficulty of

factoring a large number into 2 primes. RSA encryption and decryption require expensive

exponentiation operations. The CRT helps accelerate the operations.

In practice, RSA is used to encrypt a secret key with an encoding function.

In practice, the messages to be signed have to go through a hash function to destroy the message structures.

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Acknowledgments The notes are prepared mostly based on

D. Stinson, Cryptography: Theory and Practice, Chapman & Hall/CRC, Second Edition, 2002.

N. Ferguson and B. Schneier, Practical Cryptography, Wiley, 2003.

http://www.rsa.com/rsalabs/pkcs/files/h11300-wp-pkcs-1v2-2-rsa-cryptography-standard.pdf