Chap. 8,9: Introduction to number theory and RSA algorithm

Jen-Chang Liu, 2004

Adapted fromLecture slides by Lawrie Brown

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 labor, 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

Clifton Fadiman Clifton Fadiman was a multi-talented

writer, critic, raconteur and bookworm. He is best remembered as moderator of the erudite radio quiz show Information, Please, which ran from 1938 until 1952.

He wrote a book: The Lifetime Reading Plan ( 一生的讀書計畫 ) ，列出 西方經典一百本

Key Distribution Problem Symmetric schemes require both

parties to share a common secret key How to securely distribute this key ? often secure system failure due to a break

in the key distribution scheme

(eavesdropping)

Key Distribution methods given parties A and B have various

key distribution alternatives:1. A can select key and physically deliver to

B2. third party can select & physically deliver

key to A & B3. if A & B have communicated previously

can use previous key to encrypt a new key4. if A & B have secure communications with

a third party C, C can relay key between A & B

Not suitablefor largesystems

Initial distribution?

Scale of key distribution problem

A network with N hosts => N(N-1)/2 pairs

Node-level encryption N(N-1)/2

Application-level encryption 10 applications/node

Key distribution center (KDC)

Key distribution

center (KDC)

KDC shares a unique key (master key) with each user to distributesecret key (session key) between a pair of users: scale of key distribution problem reduces to N

EMK1 (Secret key)EMK2 (Secret key)

Secret keySecret key

Asymmetric cryptography

Real world example: locker

Public key & private key

Difference between symmetric and asymmetric cryptography

Symmetric encryption:Reverse operation is possible

Asymmetric encryption:Hard to findreverse operation

One-way trap door function

Public key

Public keyPrivate key

Motivation: RSA public-key algorithm

One key for encryption, one key for decryption

Public directory

AliceBob

Preview of RSA algorithm R: Rivest, S: Shamir, A: Adleman Plaintext M, ciphertext C

C = Me mod n, where 0 ≤ M < n

M = Cd mod n = (Med) mod n

Q: Is there e, d, n such that (Med)≡ M mod n ?A: Euler’s theorem

Q: Given e , is it possible to compute M directly from C ?

Outline Prime numbers Fermat’s and Euler’s Theorems RSA algorithm Testing for primality

Prime Numbers prime numbers only have divisors of 1 and

self they cannot be written as a product of other

numbers note: 1 is prime, but is generally not of interest

eg. 2,3,5,7 are prime, 4,6,8,9,10 are not prime numbers are central to number theory list of prime number less than 200 is:

2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199

Prime Factorisation prime factorisation: any integer a>1

can be factored in a unique way , are primes each ai is a positive

integer eg. 91=7×13 ; 3600=24×32×52

Note: factoring a number is relatively hard compared to multiplying the factors together to generate the number

tat

aa pppa 2121 tppp 21

Relatively Prime Numbers & GCD

two numbers a,b are relatively prime if have no common divisors apart from 1 eg. 8 & 15 are relatively prime since

factors of 8 are 1,2,4,8 and of 15 are 1,3,5,15 and 1 is the only common factor

conversely can determine the greatest common divisor by comparing their prime factorizations and using least powers eg. 300=22×31×52 18=21×32 hence GCD(18,300)=21×31×50=6

Outline Prime numbers Fermat’s and Euler’s Theorems RSA algorithm Testing for primality

Fermat's Theorem also known as Fermat’s Little Theorem ap-1 mod p = 1

where p is prime and gcd(a,p)=1 [a, p 互質 ]

useful in public key and primality testing

Proof of Fermat’s law Recall: p is a prime, Zp is a Galois Field

Any a multiplied by{1,2,…,p-1} will span{1,2,…,p-1} in some order

a×2a×…×((p-1)a) ≡ [(a mod p)×(2a mod p)×…×((p-1)a mod p)] mod p ≡ [1×2×…×(p-1)] mod p ≡ (p-1)! mod p

(p-1)! ap-1 ≡ (p-1)! mod p

Proof of Fermat’s law (cont.)

(p-1)! ap-1 ≡ (p-1)! mod p

Because (p-1)! is relatively prime to p

ap-1 ≡ 1 mod p

This can be re-written as

ap ≡ a mod p

Example: a = 7, p = 19 => 718 ≡ 1 mod 19 ?

718 =716×72

72=49≡11 mod 1974=72×72≡(11×11) mod 19= 7 mod 1978=74×74≡(7×7) mod 19= 11 mod 19716=78×78≡(11×11) mod 19= 7 mod 19

≡(7×11) mod 19 = 1 mod 19

Euler’s Totient Function ø(n)

when doing arithmetic modulo n complete set of residues is: 0..n-1 reduced set of residues is those numbers

(residues) which are relatively prime to n eg. for n=10, complete set of residues is {0,1,2,3,4,5,6,7,8,9} reduced set of residues is {1,3,7,9}

number of elements in reduced set of residues is called the Euler Totient Function ø(n)

i.e. is the number of positive integers less than n and relatively prime to n

Example: totient function ø(37)=?

37 is a prime, {1,…,36} are all relatively prime to 37

ø(37)=36 ø(35)=?

List them: {1,2,3,4,6,8,9,11,12,13,16,17,18,19,22,23,24,26,27,29,31,32,33,34}

ø(35)=24

Euler’s Totient Function ø(n)

Some special case: for p (p prime) : ø(p) = p-1 for p•q (pq, both prime): ø(p•q)=(p-1)×(q-1)

Proof for ø(p•q) = (p-1)×(q-1) {1,2,3,…, pq-1} 與 pq 互質的個數？

=> 與 pq 非互質個數：1. p 的倍數： {p, 2p, 3p,…, (q-1)p}

2. q 的倍數： {q, 2q, 3q,…, (p-1)q}

ø(p•q) =pq-1 – (p-1) – (q-1) = pq-p-q+1 = (p-1)(q-1)= ø(p)×ø(q)

Euler's Theorem Fermat's Theorem: an-1 ≡ 1 mod n Euler’s theorem: aø(n)≡ 1 mod n

where gcd(a,n)=1 eg.

a=3;n=10; ø(10)=4; hence 34 = 81 = 1 mod 10 a=2;n=11; ø(11)=10; hence 210 = 1024 = 1 mod 11

Proof for Euler’s theorem aø(n)≡ 1 mod n, gcd(a,n)=1 n is a prime => Fermat’s theorem Arbitrary n: ø(n) means the number of integers

that is relatively prime to n, denote the set of integers as )(21 ,,, nxxxR

Multiply each by a, modulo n:

n) mod (,,n) mod (,n) mod ( )(21 naxaxaxS

S is a permutation of R !!!* a is relatively prime to n, and xi is relative prime to n => so does axi

* There is no duplicate in S

Proof for Euler’s theorem (cont.)

)(21 ,,, nxxxR n) mod (,,n) mod (,n) mod ( )(21 naxaxaxS

is the permutation of

)(

1

)(

1

n) mod (n

ii

n

ii xax

n) (mod

)(

1

)(

1

n

ii

n

ii xax

n) (mod )(

1

)(

1

)(

n

ii

n

ii

n xxa

mod n

n) (mod 1)( na

OR n) (mod 1)( aa n

(Med)≡ M mod nRecall: RSA algorithm

Corollary to Euler’s theorem

RSA algorithm: Euler’s theorem: Given prime p and q, n=pq, 0<a<n

(Med)≡ M mod nn) (mod 1)( aa n

n) (mod 1)1)(1(1)( aaa qpn

gcd(a, n)=1

a: plaintext => not necessary prime to n !!!

1. a is prime to n => Euler’s theorem

2. a is NOT prime to n => a 是 p 的倍數 或 a 是 q 的倍數 Prove case 1: a = cp

but a qp, because 0<a<n=pq => gcd(a, q) = 1

Corollary to Euler’s theorem (cont.)

case 1: a = cp, gcd(a,q)=1Euler’s theorem

q mod 1)( qa

自乘 ø(p) 次

q mod 1 )()( pqa

=> q mod 1)( naTotient function

=> 1kq)( na

=>Multiply a=cp akcnakcpq1)( na

=> n mod 1)( aa n

Δ

Corollary to Euler’s theorem (cont.)

Given prime p and q, n=pq, 0<a<n

Alternative form of the corollary (used in RSA)

n) (mod 1)1)(1(1)( aaa qpn

n mod )( )(1)( aaa knnk

n mod )1( ak By Euler’s theorem

n mod a

RSA from Euler’s corollary RSA algorithm: find e, d, n to satisfy

Euler’s corollary: given primes p and q, n=pq,

0<a<n, and arbitrary k

(Med)≡ M mod n

n mod 1)1)(1(1)( aaa qpknk

We want ed = k( n) + 1

=> ed ≡ 1 mod ( n)

or d ≡ e-1 mod ( n)(d is the multiplicative inverse of e, it existsIf d (and e) is relatively prime to (n) )

Outline Prime numbers Fermat’s and Euler’s Theorems RSA algorithm Testing for primality

RSA algorithm – decide parameters

1. Select primes p and q.2. Calculate n=pq.3. Calculate (n)=(p-1)(q-1)4. Select e that is relative

prime to and less than (n)5. Determine d such that de ≡ 1 mod (n),and d< (n)(d is the multiplicative inverse of e, find

it using Extended Euclid’s algorithm)

Example:

p=17, q=11

n= 17x11 = 187

(n)= 16x10 = 160

e = 7

d = 23

RSA encryption/decryption example

Public key: KU={e,n}={7,187} Private key: KR={d,n}={23,187}

Ingredient and security of RSA

p, q: two primes (private, chosen)

n=pq (public, calculated)

e, with gcd((n),e)=1 (public, chosen) 1<e< (n) d ≡ e-1 mod (n) (private, calculated)

(n) must be a secret, else d can be calculated

n is known => (n) can be calculated from n?

p, q must be secrets => (n) = (p-1)(q-1)

=> p, q calculated from n?

The security of RSA Brute-force: try all private keys Mathematical attacks:

Factor n into its two prime factors, n=> p×q

Determine (n) directly Determine d directly without (n)

Timing attacks: depends on the running time of the decryption algorithm

Samecomplexity

RSA recommended key size

How to evaluate RSA security ? rough extrapolations of the running times

for smaller keys TWIRL: a hardware for integer factorization

for a 1024-bit RSA key in less than a year with only $10 million worth of hardware

Protection Lifetime of Data

Present – 2010

Present – 2030

Present – 2031 and Beyond

Minimum symmetric security level

80 bits 112 bits 128 bits

Minimum RSA key size

1024 bits 2048 bits 3072 bits

Single prime modulus RSA algorithm: Fermat’s theorem: A single prime modulus n?

(aed)≡ a mod n

an ≡ a mod n, n is a prime

ed = k(n-1) + 1

an-1 ≡ 1 mod n ak(n-1)+1 ≡ a mod n=>

ed ≡ 1 mod (n-1)=>

e, d 互為 Zn-1 下的乘法反元素However, if (e, n) is public

We can calculate d easily !!! (Zn-1 下的乘法反元素 )

MultiPrime RSA RSA: n = p q MultiPrime RSA: n = p q r ?

More primes to make up the modulus, Ex. 1024 bits RSA, 341, 341, 342 bits

primes the faster the private key operations the easier to factor

Computation – encryption/decryption

Modular exponentiation: Fast algorithm use the property:

Write exponential d as binary number: bkbk-1…b1b0

ex. 2310 = 101112 = 24+22+21+20

M = Cd mod n

(a x b) mod n = (a mod n) x (b mod n) mod n

n mod n mod n mod n mod 0

2

0

2

j

j

j

j

bb

d CCC

1123 mod 187 =(1116×114×112×111) mod 187=[(1116 mod 187)×(114 mod 187)×(112 mod 187)×(111 mod 187)] mod 187=…=88

Fast exponentiation ab mod n b=10111

1

a1

2 10

a10自乘

2+1101

a101自乘 a

2+11011

a1011自乘 a

2+110111

a10111自乘 a

ax ax =a2x

Pseudo-code for fast exponentiation

c=0; /* c will be the exponent at last */d=1; /* d will be the ab mod n at last */

for(i=k; i>=0; i--){ /* k+1 bits for b */ c = 2*c; d = (d*d) mod n; if ( bi == 1 ){ c = c+1; d = (d*a) mod n }}

Timing attacks

If this bit is 1, exec. time willbe slower

Resist to timing attacks Constant exponentiation time

return the results of exponentiation after a fixed time

Random delay Add random delay to the exp. execution

time Blinding

Multiply ciphertext by a random number

Reading assignment The code book ( 碼書 )

報告重點：年代，人物，故事，加解密方式 投影片請多用書上圖片解說故事及人物 Report one chapter per week Chap .1: 洪善群 , 張凱鈞 Chap. 2: 鄧偉華 Chap. 3: Chap. 4: Chap. 5: Chap. 6: Chap. 7: PGP 93321510, 93321538,

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