because both the system’s privacy and the security of digital money depend on encryption, a...

Because both the system’s privacy and the security of digital money depend on encryption, a breakthrough in mathematics or computer science that defeats the cryptographic system could be a disaster. The obvious mathematical breakthrough would be development of an easy way to factor large prime numbers. Any person or organization possessing this power could counterfeit money, penetrate any personal, corporate, or government file, and possibly even undermine the security of nations.

Bill Gates, The Road Ahead

David Evanshttp://www.cs.virginia.edu/~evans

CS588: Security and PrivacyUniversity of VirginiaComputer Science

Lecture 9: Security of RSA THE MAGIC WORDS ARE SQUEAMISH OSSIFRAGE.

26 Sept 2001 University of Virginia CS 588 2

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• Finding Big Pseudo Primes

• Security of RSA– Factoring


Properties of E and D

Trap-door one way function:1. D (E (M)) = M

2. E and D are easy to compute.

3. Revealing E doesn’t reveal an easy way to compute D

Trap-door one way permutation: also4. E (D (M)) = M


Property 2: Easy to Compute

• We need large “random” primes p and q

• Are there enough primes?

• How can we find them?


How many prime numbers?• Infinite (proved by Euclid, 300BC)• Proof by contradiction:

Suppose that there exist only finitely many primes p1 < p2 < ... < pr.

Let N = (p1)(p2)...(pr) + 1

N > pr so it is composite, N = p * M

If p = pi for some 1…r, then,

N = pi * M = pi * (p1)(p2)...(pi-1) (pi+1)...(pr) + 1

pi (M - (p1)(p2)...(pi-1) (pi+1)...(pr)) = 1

Contradiction: pi > 1

Hence, there must be infinitely many primes.


Density of Primes

From http://www.utm.edu/research/primes/howmany.shtml

(x) is the number of primes x


Approximating (x) • The Prime Number Theorem:

(x) ~ x/ln x – Difficult to prove (first conjectured by Legendre

in 1798 by looking at table of values)

• How many guesses to find a prime bigger than x?– About ln x/2 guesses

• (Naïvely) Each guess requires sqrt(x) work• For 200 digits: 230 guesses * 10100

– More work than breaking 3DES!


Need a faster prime test

• There are several fast probabilistic prime tests

• Can quickly test a prime with high probability, with a small amount of work

• If we pick a non-prime, its not a disaster (exercise for reader, will be on PS3)


Fermat Test• Recall Fermat’s Little Theorem:

If n is prime and a is not divisible by n then

an-1 1 mod n• Prove n is composite by finding an-1 1 mod n• Showing an-1 1 mod n does not prove it is prime• But if it holds for many a’s it is likely than n is

prime – Holds for all a’s for some non-primes known as

Carmichael Numbers: 561, 645, 1105, …)

• Better prime test: Miller-Rabin – Probability n is prime 1 – ¼k


Property 4: E (D (M)) = M

D(M) = Md mod n

E(D(M)) = (Md mod n)e mod n

= Mde mod n

= Med mod n

= M

(from the property 1 proof)







Are there other functions that have properties 1, 2 and 4?


RSA

E(M) = Me mod n

D(C) = Cd mod n

n = pq p, q are prime

d is relatively prime to (p – 1)(q – 1)

ed 1 (mod (p – 1)(q – 1))


Revealing E doesn’t reveal D• Revealing E: e, n.

• Can attacker find D?

• If attacker factors n = p * qed 1 mod (p – 1)(q – 1)

Easy to find d e-1 mod (p – 1)(q – 1)

• Use experience to argue factoring is hard.

• Argue all other attacks are at least as hard as factoring n.


Gardner’s Column: Original RSA challenge ($100)

n (RSA-129) = 1 1438 1625 7578 8886 7669 2357 7997 6146 6120 1021 8296 7212 4236 2562 5618 4293 5706 9352 4573 3897 8305 9712 3563 9587 0505 8989 0751 4759 9290 0268 7954 3541

e = 9007

C = 9686 9613 7546 2206 1477 1409 2225 4355 8829 0575 9991 1245 7431 9874 6951 2093 0816 2982 2514 5708 3569 3147 6622 8839 8962 8013 3919 9055 1829 9451 5781 5154

Scientific American, August 1977


40000000000000000 17

Ron Rivest (1977): factoring n (129 digits) would require at least 40 quadrillion years if you could do a * b mod c in one nanosecond.

Derek Atkins (April 1994): We are happy to announce that RSA-129 = 3490 5295 1084 7650 9491 4784 9619 9038 9813 3417 7646 3849 3387 8439 9082 0577 * 3 2769 1329 9326 6709 5499 6198 8190 8344 6141 3177 6429 6799 2942 5397 9828 8533


Trial and Error Factoring• Guess x, if 1 < gcd (x, n) < n then x is an

interesting factor• If p and q are similar size, lowest factor is

around n.

– Requires O(n) divisions.– For RSA-129 = 1.1 * 1064 divisions, 1 per

nanosecond = 3.4 * 1047 years


Pollard’s Rho Method

• Fastest known in 1977 [Pollard75]

• To find factor p, requires 4p modular multiplies

• Worst case: lowest p is n, we need 4n multiplies

• For RSA-129 = 1.3 * 1032 = 4 * 1015 years

• Rivest probably used this, but made a math error (4 quadrilllion 40 quadrilllion)


How so Fast• Better factoring algorithms• Distributed computation

• Still can’t do a * b mod c in one nanosecond (not faster processors)– 1ns = 10-9 s– Best Pentium 4 today: 2 GHz (cycle = .5ns)– But, multiplying 100 digit numbers takes many

cycles


Fermat Factoring

• Factor 8051• Hint: (½ (a + b))2 – (½(a – b))2

= ¼ (a2 + 2ab + b2) - ¼ (a2 - 2ab + b2) = ½ ab + ½ ab = ab

• 8051 = 8100 – 49 = 902 – 72

• 8051 = 97 83 ½ (83 + 97) = 90½ (83 – 97) = 7


Fermat Factoring

• Any odd composite integer can be written as n = u2 - v2

• To factor n, start looking for u and v around n

• Is this less work than trial and error division? No, only in lucky cases

where factor is near n


Kraitchik’s Enhancement

• Instead of looking for u and v such that n = u2

- v2 look for any multiple of n:n = (u2 - v2)k

or: u2 v2 mod n

• Some solutions uninteresting: u v mod n

• Others (½) are interesting: u v mod nn divides u2 - v2 = (u – v) (u + v) but not (u – v) or (u + v)

gcd (u – v, n) and gcd (u + v, n) are factors of n


Kraitchik, cont.For x’s near n

Q(x) = x2 – nTry to find x’s where:

Q(x1) Q(x2) …Q(xk) = v2

x1x2 …xk = uThen,

x12x2

2 …xk2 = u2 (x1

2 – n) … (xk2 – n) mod n

v2 mod nBut, how do we find the x’s?


Finding the x’s• Some of the Q(x)’s are easy to factor• From these, find products that are squares

n = 2041 462 = 2116Q(46) = 75 = 3 52

Q(47) = 168 = 23 3 7Q(48) = 263 (hard to factor)Q(49) = 360 = 23 32 5Q(50) = 459 = 33 17Q(51) = 560 = 24 5 7Q(46) Q(47) Q(49) Q(51) = 23+3+4 34 54 72

= (25 32 52 7)2

Example from [Pomerance96]

Looking for:

Q(x1) Q(x2) …Q(xk) = v2

x1x2 …xk = u


Finding the FactorsQ(46) Q(47) Q(49) Q(51) = (25 32 52 7)2 v2

u = 46 47 49 51 = 311 mod 2041v = 25 32 52 7 = 1416 mod 2041

u2 = v2 mod 2041

311 1416 mod 2041So, gcd (1416 – 311, 2041) is a factor of 2041.

gcd (1105, 2041) = gcd (2041, 1105) = gcd (2041, 1105 mod 2041)= gcd (1105, 2041 mod 1105) = gcd (1105, 936)= gcd (936, 1105 mod 936) = gcd (936, 169)= gcd (169, 936 mod 169) = gcd (169, 91)= gcd (91, 169 mod 91) = gcd (91, 78)= gcd (78, 91 mod 78) = gcd (78, 13)

= gcd (13, 78 mod 13) = gcd (13, 13) = 13


Factoring Pragmatics

• Conjectured to take esqrt(2 log n log log n) steps to factor n.

• Quadratic Sieve Factoring– Improved Kraitchik’s algorithm to choose

likely good values for x’s– Complexity: esqrt(log n log log n)

– How much bigger numbers can it factor?

Twice as many digits in same time!


Breaking RSA-129

• Organized by Derek Atkins and others, 1994• Quadratic Sieve algorithm• Memory-limited (1994 – most workstations

16MB RAM), used 10M to hold .5M primes• Recruited volunteers from Internet• 1600 machines• Used 5000 MIPS years over 8 months


To factor r [RSA-129] we assembled the largest collaboration yet seen in computational number theory and, possibly, performed the largest single computation ever completed. In several important respects, the resources we had available were barely adequate for the task. Consequently, ingenuity and diplomacy were required for the successful completion of the project.

Derek Atkins, Michael Graff, et. al., The Magic Words Are Squeamish Ossifrage,

AsiaCrypt 1994.


Recent Factoring Algorithms

• Team from CWI (Amsterdam) factored RSA-155 (512 bits), August 1999

• Number Field Sieve (Sneakers)

• ~8000 MIPS years (36 CPU years)

• 7 months on ~300 machines

• Lecture 1: factor 300 digit number for automatic A. How much harder is this?


RSA Security

• Factoring is hard (except in Sneakers)

• Is this enough to know RSA is secure?

• Can you compute D without factoring n?– Probably not, but can’t prove it.– But, can prove other obvious mathematical

attacks are equivalent to factoring.


(n) without factoring• Calculate (n) without factoring n.

ed 1 mod (n)• Equivalent to factoring:

(n) = (p – 1)(q – 1) = n – (p + q) + 1 p + q = n – (n) – 1 p – q = sqrt ((p + q)2 – 4n) [next slide]

2p = sqrt ((n – (n) –1)2 – 4n) + n – (n) –1If we know (n) we can calculate p and q easily.


p – q = sqrt ((p + q)2 – 4n)

(p – q)2 = p2 – 2pq + q2

= p2 + q2 – 2n

(p + q)2 = p2 + q2 + 2n

(p – q)2 = (p + q)2 – 4n

p – q = sqrt ((p + q)2 – 4n)


Determine d without (n) • Brute force: m digits long, amount of

work is 10m. – Try 1T/second (special purpose hardware)– Will take 3*1030 years for m = 50.– For factoring difficulty, m > 100. (Bigger

than AES key)

• Non-brute force: knowing d enables factoring.


Determining d factoring

ed = 1 mod (n)k * (n) = ed – 1

Already showed, finding (n) is same as factoring.

Also true for multiple of (n).


Properties of RSA’s E and D

• Trap-door one way function:D (E (M)) = ME and D are easy to compute.Revealing E doesn’t reveal an easy way to

compute D

• Trap-door one way permutation: also E (D (M)) = M


Applications of RSA• Privacy:

– Bob encrypts message to Alice using EA

– Only Alice knows DA

• Signatures: – Alice encrypts a message to Alice using DA

– Bob decrypts using EA

– Knows it was from Alice, since only Alice knows DA

• Things you use every day: ssh, SSL, DNS, etc.• More than 400,000,000 copies of the RSA

algorithm are currently installed


Two “Questionable” Statements in RSA Paper“(The NBS scheme (DES) is probably somewhat faster if special-purposed hardware encryption devices are used; our scheme may be faster on a general-purpose computer since multiprecision arithmetic operations are simpler to implement than complicated bit manipulations.)”

(p. 4)


Two “Questionable” Statements in RSA Paper

“The need for a courier between every pair of users has thus been replaced by the requirement for a single secure meeting between each user and the public file manager when the user joins the system.”

(p. 6)


Who really invented RSA?• General Communications HQ, Cheltenham

(formed from Bletchley Park after WWII)• 1969 – James Ellis asked to work on key

distribution problem• Secure telephone conversations by adding

“noise” to line:– Receiver could add the noise and subtract it– No need for sender to have key!

• Late 1969 – idea for PK, but no function• Clifford Cocks invented function (RSA)


Charge• Next time: what to do about all those

couriers!• Project Pre-Proposals due Monday• They should include:

– Clear topic description and plan: the more details you can provide on what you plan to do, the better feedback I can provide

– Related Work• Summary of what you have found so far• Reading list: things you plan to read next

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