Cryptography and Network Security - Åbo Akademiusers.abo.fi/ipetre/crypto/lecture7.pdf ·...

January 31, 2012 1

Cryptography and Network Security

Lecture 7: Public-key cryptography and RSA Ion Petre Department of IT, Åbo Akademi University

Spring 2012 http://users.abo.fi/ipetre/crypto/

http://users.abo.fi/ipetre/crypto/

http://www.abo.fi/~ipetre/crypto/�



January 31, 2012 2

Some unanswered questions on symmetric cryptosystems

Key management: changing the secret key or establishing one is nontrivial Change the keys two users share (should be done reasonably often) Establish a secret key with somebody you do not know and cannot meet

in person: (e.g., visiting secure websites such as e-shops) This could be done via a trusted Key Distribution Center (details in a

future lecture) Can (or should) we really trust the KDC? “What good would it do after all to develop impenetrable cryptosystems, if

their users were forced to share their keys with a KDC that could be compromised by either burglary or subpoena?” – Diffie, 1988

Digital signatures


January 31, 2012 3

A breakthrough idea

Rather than having a secret key that the two users must share, each users has two keys One key is secret and he is the only one who knows it The other key is public and anyone who wishes to send him a

message uses that key to encrypt the message Diffie and Hellman first (publicly) introduced the idea in 1976 – this was

radically different than all previous efforts NSA claims to have known it since mid-1960s! Communications-Electronic Security Group (the British counterpart of

NSA) documented the idea in a classified report in 1970


January 31, 2012 4

A word of warning

Public-key cryptography complements rather than replaces symmetric cryptography There is nothing in principle to make public-key crypto more secure than

symmetric crypto Public-key crypto does not make symmetric crypto obsolete: it has its

advantages but also its (major) drawbacks such as speed Due to its low speed, it is mostly confined to key management and

digital signatures


January 31, 2012 5

The idea of public-key cryptography

The concept was proposed in 1976 by Diffie and Hellman although no practical way to design such a system was suggested

Each user has two keys: one encryption key that he makes public and one decryption key that he keeps secret Clearly, it should be computationally infeasible to determine the

decryption key given only the encryption key and the cryptographic algorithm

Some algorithms (such as RSA) satisfy also the following useful characteristic: Either one of the two keys can be used for encryption – the other one

should then be used to decrypt the message First we will investigate the concept with no reference yet to practical

design of a public-key system


January 31, 2012 6

Essential steps in public-key encryption

Each user generates a pair of keys to be used for encryption and decryption

Each user places one of the two keys in a public register and the other key is kept private

If B wants to send a confidential message to A, B encrypts the message using A’s public key

When A receives the message, she decrypts it using her private key Nobody else can decrypt the message because that can only be done

using A’s private key Deducing a private key should be infeasible

If a user wishes to change his keys – generate another pair of keys and publish the public one: no interaction with other users is needed


January 31, 2012 7

Bob sends an encrypted message to Alice


January 31, 2012 8

Some notation

The public key of user A will be denoted KUA

The private key of user A will be denoted KRA

Encryption method will be a function E Decryption method will be a function D If B wishes to send a plain message X to A, then he sends the

cryptotext Y=E(KUA,X) The intended receiver A will decrypt the message: D(KRA,Y)=X


January 31, 2012 9

A first attack on the public-key scheme – authenticity

Immediate attack on this scheme: An attacker may impersonate user B: he sends a message E(KUA,X)

and claims in the message to be B – A has no guarantee this is so This was guaranteed in classical cryptosystems simply through knowing

the key (only A and B are supposed to know the symmetric key) The authenticity of user B can be established as follows:

B will encrypt the message using his private key: Y=E(KRB,X) This shows the authenticity of the sender because (supposedly) he is the

only one who knows the private key The entire encrypted message serves as a digital signature Note: this may not be the best possible solution: ideally, digital signatures

should be rather small so that one can preserve many of them over a long period of time Better schemes will be presented a couple of lectures on


January 31, 2012 10

A scheme to authenticate the sender of the message


January 31, 2012 11

Encryption and authenticity

Still a drawback: the scheme on the previous slide authenticate but does not ensure security: anybody can decrypt the message using B’s public key

One can provide both authentication and confidentiality using the public-key scheme twice: B encrypts X with his private key: Y=E(KRB,X) B encrypts Y with A’s public key: Z=E(KUA,Y) A will decrypt Z (and she is the only one capable of doing it):

Y=D(KRA,Z) A can now get the plaintext and ensure that it comes from B (he is the

only one who knows his private key): decrypt Y using B’s public key: X=E(KUB,Y)


January 31, 2012 12

Secrecy and authentication using public-key schemes


January 31, 2012 13

Applications for public-key cryptosystems

1. Encryption/decryption: sender encrypts the message with the receiver’s public key

2. Digital signature: sender “signs” the message (or a representative part of the message) using his private key

3. Key exchange: two sides cooperate to exchange a secret key for later use in a secret-key cryptosystem


January 31, 2012 14

Requirements for public-key cryptosystems

Generating a key pair (public key, private key) is computationally easy

Encrypting a message using a known key (his own private or somebody else’s public) is computationally easy

Decrypting a message using a known key (his own private or somebody else’s public) is computationally easy

Knowing the public key, it is computationally infeasible for an opponent to deduce the private key

Knowing the public key and a ciphertext, it is computationally infeasible for an opponent to deduce the private key

Useful extra feature: encryption and decryption can be applied in any order:

E( KUA, D(KRA,X) ) =D(KRA, E( KUA, X) )


January 31, 2012 15

Designing a public-key cryptosystem

Computationally easy usually means polynomial-time algorithm Computationally infeasible more difficult to define

Usually means super-polynomial-time algorithms, e.g., exponential-time algorithms Classical complexity analysis (worst-case complexity or average-case complexity) are

worthless in cryptography – one should make sure a problem is difficult for virtually all inputs and not just in the worse or in the average case

Public-key cryptosystems usually rely on difficult math functions rather than S-P networks as classical cryptosystems One-way function: easy to calculate in one direction, infeasible to calculate in the

other direction (i.e., the inverse is infeasible to compute) Trap-door function: difficult function that becomes easy if some extra information is

known Aim: find a trap-door one-way function for encryption – decryption will be the inverse


January 31, 2012 16

RSA

One of the first proposals on implementing the concept of public-key cryptography was that of Rivest, Shamir, Adleman – 1977: RSA

The RSA scheme is a block cipher in which the plaintext and the ciphertext are integers between 0 and n-1 for some fixed n Typical size for n is 1024 bits (or 309 decimal digits) To be secure with today’s technology size should between 1024 and 2048 bits

Idea of RSA: it is a difficult math problem to factorize (large) integers Choose p and q odd primes, n=pq Choose integers d,e such that Med=M mod n, for all M<n Plaintext: block of k bits, where 2k<n≤2k+1 – can be considered a number M with

M<n Encryption: C=Me mod n Decryption: Cd mod n = Mde mod n = M Public key: KU={e,n} Private key: KR={d,n}

Questions: How do we find d,e? How do we find large primes? Answer: Number Theory!


January 31, 2012 17

Motto for our introduction to Number Theory

The Devil said to Daniel Webster: "Set me a task I can't carry out, and I'll give you anything in the world you ask for."

Daniel Webster: "Fair enough. Prove that for n greater than 2, the equation an + bn = cn has

no non-trivial solution in the integers." They agreed on a three-day period for the labour, and the Devil disappeared. At the end of three days, the Devil presented himself, haggard, jumpy, biting his lip. Daniel

Webster said to him, "Well, how did you do at my task? Did you prove the theorem?' "Eh? No . . . no, I haven't proved it." "Then I can have whatever I ask for? Money? The Presidency?' "What? Oh, that—of course. But listen! If we could just prove the following two lemmas—" —The Mathematical Magpie, Clifton Fadiman


January 31, 2012 18

Notions of number theory

Fermat’s little theorem: if p is prime and a is positive integer not divisible by p, then ap-1 ≡ 1 mod p

Corollary: For any positive integer a and prime p, ap ≡ a mod p Comments:

This is a first step in our quest to find Med=M mod n – not quite enough though Fermat’s little theorem provides a necessary condition for an integer p to be

prime – the condition is not sufficient We will turn this theorem into a (probabilistic) test for primality

Example: p=5, a=3, 35=243=3 mod 5 p=5, a=10, 105=100000=10 mod 5 = 0 mod 5

Fermat’s theorem, as useful as it will turn out to be, it does not provide us with integers d,e we are looking for – Euler’s theorem (a refinement of Fermat’s) does


January 31, 2012 19

Euler’s totient function

Euler’s function associates to any positive integer n a number φ(n): the number of positive integers smaller than n and relatively prime to n Example:

φ(37)=36 φ(p)=p-1, for any prime p φ(35)=24: {1,2,3,4,6,8,9,11,12,13,16,17,18,19,22,23,24,26,27,29,31,32,33,34}

Easy to see that for any two distinct primes p,q, φ(pq)=(p-1)(q-1) All numbers smaller than pq are relatively primes with pq except for multiples of

p (q-1 of them) and multiples of q (p-1 of them): pq-(q-1)-(p-1)=(p-1)(q-1) Euler’s theorem: for any relatively prime integers a,n we have

aφ(n)≡1 mod n Corollary: For any integers a,n we have aφ(n)+1≡ a mod n Corollary: Let p,q be two odd primes and n=pq. Then:

φ(n)=(p-1)(q-1) For any integer m with 0<m<n, m(p-1)(q-1)+1≡m mod n For any integers k,m with 0<m<n, mk(p-1)(q-1)+1≡m mod n


January 31, 2012 20

Back to RSA

Euler’s theorem provides us the numbers d,e such that Med=M mod n We have to choose d,e such that ed=kφ(n)+1 for some k Equivalently, d≡e-1 mod φ(n) To calculate the modular inverse of an interger: the extended Euclid’s algorithm!

see Lecture 5 The RSA scheme

Key generation Choose two odd primes p,q – keep private. Compute n=pq – make public Choose e, 1<e<φ(n) with gcd(φ(n),e)=1 – make public Compute d≡e-1 mod φ(n) – keep private Private key is {d,n} Public key is {e,n}

Encryption Plaintext: block of k bits, where 2k<n≤2k+1 – can be considered a number M with M<n Ciphertext: C=Me mod n

Decryption: Ciphertext: C Plaintext: Cd mod n = Mde mod n = M


January 31, 2012 21

Example

Key generation Select primes p=17, q=11 Compute n=pq=187 Compute φ(n)=(p-1)(q-1)=160 Select e=7 Compute d: d=23 (use the extended

Euclid’s algorithm) KU={7,187} KR={23,187}

Encrypt M=88: 887 mod 187 887 mod 187 = [ (884 mod 187)(882 mod

187) (88 mod 187) ] = 11 Decrypt C=11: 1123 mod 187

M=1123 mod 187= [ (1116 mod 187)(114 mod 187) (112 mod 187)(11 mod 187)]

112 mod 187 =121 114 mod 187= 1212 mod 187=55 118 mod 187=552 mod 187= 33 1116 mod 187=332 mod 187=154 M=154 x 55 x 121 x 11 mod 187 = 88

RSA scheme –Key generation

•Choose primes p,q •Compute n=pq •Choose e, 1<e<φ(n) with gcd(φ(n),e)=1 •Compute d≡e-1 mod φ(n) •Private key is {d,n} •Public key is {e,n}

–Encryption •C=Me mod n

–Decryption: •Cd mod n = Mde mod n = M


January 31, 2012 22

Computational aspects – RSA implementation Fast modular exponentiation

Take each step in turn and discuss how can it be implemented efficiently

For encryption and decryption we must be able to do quick modular exponentiations – two ideas are useful: (ab mod n) = [(a mod n)(b mod n)] (mod n) To compute x16 mod n we do not have to

do 15 multiplication but only 4: compute x2 mod n, x4 mod n, x8 mod n, x16 mod n

Apply this to compute quickly any exponent, not just powers of 2






January 31, 2012 23

Fast modular exponentiation

Square-and-multiply algorithm Input: n,x,b (b is in base 2 (bk-1,…,b1,b0), b≠0 Output: xb mod n 1. z=1 2. for i=k-1 downto 0

3. z=z2 mod n 4. if bi=1 then z=zx mod n

Complexity O(r3), where r=[log2n] Example: encrypt 9726 with

KU={3533,11413}: 97263533mod 11413 3533=(1,1,0,1,1,1,0,0,1,1,0,1) Ciphertext: 5761





i bi z

11 1 9726

10 1 97262x9726=2659

9 0 26592=5634

8 1 56342x9726=9167

7 1 91672x9726=4958

6 1 49582x9726=7783

i bi z

5 0 77832=6298

4 0 62982=4629

3 1 46292x9726=10185

2 1 101852x9726=105

1 0 1052=11025

0 1 110252x9726=5761


January 31, 2012 24

Computational aspects – RSA implementation Key generation

The highlighted part in the algorithm is easy to implement Generate a series of random numbers

and test each against φ(n) for relative primality

Testing whether or not two integers are relatively prime and finding a modular inverse can be done with the extended Euclid’s algorithm

Very few tests are needed before a usable e is found: the probability that two random numbers are relatively prime is 0.6






January 31, 2012 25


No practical techniques to yield large prime numbers

Procedure: generate random odd numbers and test whether that integer is prime

Testing whether or not an integer n is a prime is a difficult problem (“primality is difficult”) There has been a long standing question

in math whether or not primality can be tested in polynomial deterministic time

Answer (2002): YES! Manindra Agrawal, Neeraj Kayal and Nitin

Saxena, “PRIMES is in P”, Ann. of Math. (2), 160:2 (2004) 781--793.

Drawback: high complexity – O(log12n f(log log n)), where f is a polynomial






January 31, 2012 26

Miller-Rabin primality test

Faster methods of testing primality exist – they are all probabilistic Such an algorithm can give two answers to the question “Is n prime?”

1. No, it is not 2. n is probably prime The probability can be made arbitrarily large

Other algorithms may give precise answer but with low probability they may take a long time to finish

Most popular primality test: Miller- Rabin, based on Fermat’s little theorem


January 31, 2012 27


Fermat’s little theorem: if p is prime and a is positive integer not divisible by p, then ap-1 ≡ 1 mod p

Idea of the Miller-Rabin test: We need to test if the odd integer n is prime: test the equality in Fermat’s little theorem

for n and a random a A speedup may be done so that we do not have to compute all powers of a – details bellow

n-1 is even, i.e., of the form n-1=2kq, with k>0, q odd: k and q easy to find Choose an integer a such that 1<a<n-1 Compute modulo n the values a2jq, 0≤j≤q: aq, a2q,…, a2k-1q, a2kq

By Fermat’s theorem, if n is prime, then the last value in the sequence is 1 – the sequence may have some other 1s, consider the first 1 in the sequence Case 1: the first number in the sequence is 1 – then all other powers are also 1 Case 2: some number a2jq in the sequence is 1 – in this case a2j-1q = n-1 mod n

0 = (a2jq -1) mod n = (a2j-1q – 1) (a2j-1q + 1) mod n, i.e., n divides (a2j-1q – 1) or (a2j-1q + 1) Since we took the first 1 in the sequence, it follows that n divides (a2j-1q + 1): a2j-1q = n-1 mod n

The test: if either the first element in the sequence is 1, or some other element is n-1, then n could be prime. Otherwise n is certainly not prime


January 31, 2012 28


TEST(n) 1. n-1=2kq: compute k and q 2. Select a random integer a,

1<a<n-1 3. If aqmod n=1 then return

“probably prime” 4. For j=0 to k-1 do

5. If a2jq mod n = n-1, then return “probably prime”

6. Return “not a prime”

•Question: for how many integers a does the test fail? •Failure: n is not prime but the algorithm return “probably prime”

•Answer: for at most (n-1)/4 integers a with 1≤a≤n-1 •Thus, the probability of failure is at most ¼

•Practical implementation: •Repeatedly invoke TEST(n) using random choices for a •If TEST(n) return at least once “not a prime”, then n is not a prime •If t executions of TEST(n) return “probably prime”, then the probability that n is indeed a prime is larger than 1-4-t

•t=10 gives probability larger than 0.999999


January 31, 2012 29


To choose primes p,q we generate random numbers p,q on the desired scale of magnitude and test the primality with Miller-Rabin

Question: How many trials should we expect to do before we find a prime?

Distribution of primes Prime number theorem: for any integer x, the

primes near x are spaced in average one every log(x) integers

On average we have to test log(x) integers before we find a prime – reject immediately even integers and integers ending in 5

Correct rate: we need to test in average 0.4 log(x) integers before we find a prime of the order of x

Example: if we look for a prime on the order of magnitude 2200 we need to do in average 55 trials, order of magnitude 21024: in average 284 trials






January 31, 2012 30

Attacking RSA Brute force attacks: try all possible private keys

As in the other cases defend using large keys: nowadays integers between 1024 and 2048 bits

Mathematical attacks Factor n into its two primes p,q: this is a hard problem

for large n Challenges by RSA Labs to factorize large integers Last solved challenge: 768 bits (2009)

Determine φ(n) directly without first determining p,q: this math problem is equivalent to factoring

Determine d directly, without first determining φ(n): this is believed to be at least as difficult as factoring

Suggestions for design The larger the keys, the better but also the slower the

algorithm Choosing p,q badly may weaken the algorithm

p,q should differ in length by only a few bits: for a 1024-bit key, p,q should be on the order of magnitude 1075 to 10100

p-1 and q-1 should both contain a large prime factor gcd(p-1,q-1) should be small d should be larger than n1/4


•Choose primes p,q •Compute n=pq •Choose e with gcd(φ(n),e)=1 •Compute d≡e-1 mod φ(n) •Private key is {d,n} •Public key is {e,n}




January 31, 2012 31

Attacks on RSA Timing attacks: determine a private key by keeping track

of how long a computer takes to decipher a message (ciphertext-only attack) – this is essentially an attack on the fast exponentiation algorithm but can be adapted for any other algorithm Whenever a bit is 1 the algorithm has more computations to

do and takes more time Countermeasures:

Ensure that all exponentiations take the same time before returning a result: degrade performance of the algorithm

Add some random delay: if there is not enough noise the attack succeeds

Blinding: multiply the ciphertext by a random number before performing exponentiation – in this way the attacker does not know the input to the exponentiation algorithm. (implemented in the commercial products from RSA Data Security Inc.) Decryption M=Cd mod n is modified as follows: Generate a secret random number r between 0 and n-1 Compute C’=C(re) mod n where e is the public exponent Compute M’=(C’d) mod n with the ordinary exponentiation Compute M=M’r-1 mod n Reported performance penalty: 2 to 10%

Square-and-multiply algorithm – Input: n,x,b

(b is in base 2 (bk-1 ,…,b1,b0) – Output: xb mod n 1. z=1 2. for i=k-1 downto 0

3. z=z2 mod n 4. if bi=1 then z=zx mod n


January 31, 2012 32

Pseudo-random number generators

Essential in RSA (and elsewhere) to be able to generate pseudo-random numbers

A sequence of numbers is random if they have uniform distribution and are independent (no value can be deduced from the others)

We generally use algorithmic techniques to generate such numbers – they will not be independent and thus not random The whole point is to make them “look” random, i.e., make them pass many

test of randomness Three tests to be used in evaluating a pseudo-random number

generator The function should be full-period generating function: generate all numbers

in its range before repeating The generated sequence should appear random: pass many statistical tests The function should implement efficiently with 32-bit arithmetic


January 31, 2012 33

Pseudo-random number generators

The most widely used technique is the linear congruential method (Lehmer 1951)

Xn+1=(aXn+c) mod m One should be very careful in choosing constants a, c, m: a=c=1 is bad

choice! Value of m should be as large as possible: usually close to 231, very often

chosen to be the prime number 231-1; in this case one can take c=0 There are very few good choices for a: for m= 231-1 only a handful of choices

are advisable – very often used is a=75=16807 Xn+1=16807 Xn mod (231-1)

Using this in cryptography needs extra care: If the attacker finds one single value, then he will be able to compute all

subsequent values Idea: restart the sequence often, using the clock as seed (initial value)


January 31, 2012 34

Cryptographically generated pseudo-random numbers

Idea: use cryptographic primitives to generate pseudo-random numbers

One possibility: Use a counter and encrypt each value for the counter, e.g., with DES – the cryptotext will be the key Stronger version: instead of a counter use a

PRNG (pseudo-random number generator) Technique can be made stronger using a more

sophisticated scheme and 3DES, see ANSI X9.17 PRNG


January 31, 2012 35

Another speed-up in RSA implementation

Operations modulo big integers become more time-consuming as the integers grows bigger

Efficient implementation: use Chinese Remainder Theorem (CRT) In its simplest formulation, CRT essentially says that if n=pq, then

instead of addition/difference/multiplication modulo n one can perform the same modulo p and modulo q and then compute the result mod n

Big advantage because the modules are much smaller


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