modular arithmetic

Modular Arithmetic

p-1

´p 1

n|m means that m is a an integer multiple of n.

We say that “n divides m”.

7|21 7 divides 21

2|4 2 divides 47|21 7 divides 21

(a mod n) means the remainder when a is

divided by n.

If ad + r = n, 0· r < nThen r = (a mod n)and d = (a div n)

Egyptian Division

95 15 80* 40 20 10*

8 4 2 1

Double and Subtract

95/10 = 8+1 remainder 5

Egyptian in the exponent: Calculate ab mod n:

Except for b, work in a reduced mod system to keep all intermediate results less than b log2 (n) c + 1 bits long.

Phase I (Repeated Multiplication)

For log b stepsmultiply largest so far by a(a, a2, a4, … )

Phase II (Make ab from bits and pieces)Expand n in binary to see how n is the

sum powers of 2. Assemble ab by multiplying together appropriate powers of a.

Modular equivalence of integers a and b:

a ´ b [mod n]a ´n b

“a and b are equivalent modulo n”

iff (a mod n) = (b mod n)iff n|(a-b)

Equivalence mod n implies equivalence mod any divisor of

n.

If (x ´n y) and (k|n)Then: x ´k y

Fundamental lemma of plus, minus, and times modulo n:

If (x ń y) and (a ń b)Then: x+a ń y+b

x-a ń y-b xa ń yb

Fundamental lemma of plus minus, and times modulo n:

When doing plus, minus, and times modulo n, I can at any

time in the calculation replace a number with a number in the same residue class modulo n

Please calculate in your head:

329 * 666 mod 331-2 * 4 = -8 = 323

The reduced system modulo

n:

Zn = {0, 1, 2, …, n-1}

Define +n and *n:a +n b = (a+b mod n)

a *n b = (a*b mod n)

The reduced system modulo 3

Z3 = {0, 1, 2}

Two binary, associative operators on Z3:+3

0 1 2

0 0 1 2

1 1 2 0

2 2 0 1

*3 0 1 2

0 0 0 0

1 0 1 2

2 0 2 1

The reduced system modulo 2

Z2 = {0, 1}

Two binary, associative operators on Z2:

+2

0 1

0 0 1

1 1 0

*2 0 1

0 0 0

1 0 1

The Boolean interpretation ofZ2 = {0, 1}

0 means FALSE 1 means TRUE

+2

XOR

0 1

0 0 1

1 1 0

*2

AND

0 1

0 0 0

1 0 1

The reduced systemZ4 = {0, 1,2,3}

+ 0 1 2 3

0 0 1 2 3

1 1 2 3 0

2 2 3 0 1

3 3 0 1 2

* 0 1 2 3

0 0 0 0 0

1 0 1 2 3

2 0 2 0 2

3 0 3 2 1

The reduced systemZ5 = {0, 1,2,3,4,5}

+ 0 1 2 3 4

0 0 1 2 3 4

1 1 2 3 4 0

2 2 3 4 0 1

3 3 4 0 1 2

4 4 0 1 2 3

* 0 1 2 3 4

0 0 0 0 0 0

1 0 1 2 3 4

2 0 2 4 1 3

3 0 3 1 4 2

4 0 4 3 2 1

Theorem: There are exactly n/(c,n) distinct multiples of c modulo n.

Example: n=8, c=3.

7

5 3

1

0

6 2

4


Example: n=8, c=4.

7

5 3

1

0

6 2

4


Example: n=8, c=8.

7

5 3

1

0

6 2

4


Example: n=8, c=6.

7

5 3

1

0

6 2

4

The multiples of c modulo n is the set:{0, c, c +n c, c +n c +n c, ….}

= {kc mod n | 0· k · n-1}

7

5 3

1

0

6 2

4

Multiples of 6

Theorem: There are exactly n/(c.n) distinct multiples of c modulo n:

{kc mod n | 0· k < n/(c,n)}

Let k= n/(c,n), which implies ck ´n 0There are · k distinct multiples of c mod

n: c*0, c*1, c*2, …, c*(k-1)

cx ´n cy $ n|c(x-y) $ (x-y) ¸ n/(c.n) = kThere are ¸ k multiples of c

Is there a fundamental lemma of division modulo n?

cx ´n cy ) x ´n y ?

2*2 ´6 2*5, but not 2 ´6 5.

Repaired fundamental lemma of division modulo n?

c 0 (mod n), cx ´n cy ) x ´n y ?

6*3 ´10 6*8, but not 3 ´10 8.

Which c are safe for cancellation mod n, i.e., cx ´n cy

) x ´n y ?

Hint: There are only n/(n,c) distinct multiples of c.

Which c are safe for cancellation mod n, i.e., cx ´n cy

) x ´n y ?

Hint: There are only n/(n,c) distinct multiples of c.

If (c,n)\neq 1, then there exists distinct x,y<n s.t. cx=cy.

Fundamental lemma of division modulo n.

(c,n)=1, ca ´n cb ) a ´n b

mod

| ( )

| ( )

| since ( , ) 1

mod

ab ac n

n ab ac

n a b c

n b c a n

b c n

A sense in which you can always cancel by c:

Theorem: cx ń cy ) x ń/(c,n) y

cx ń cy ) cx ń/(c,n) cy and ( c, n/(c,n) )=1

) x ń/(c,n) y

Fundamental lemma of division modulo n.

(c,n)=1, ca ´n cb ) a ´n b

Zn* = {x 2 Zn | GCD(x,n) =1}

Multiplication over Zn* will have the

cancellation property.

Z12* = {1,5,7,11}

*12 1 5 7 11

1 1 5 7 11

5 5 1 11 7

7 7 11 1 5

11 11 7 5 1

Z15*

* 1 2 4 7 811

13

14

1 1 2 4 7 811

13

14

2 2 4 814

1 711

13

4 4 8 113

214

711

7 714

13

4 11 2 1 8

8 8 1 211

413

14

7

11

11

714

2 13 1 8 4

13

13

11

7 1 14 8 4 2

14

14

13

11

8 7 4 2 1

The column permutation property is equivalent to the right cancellation

property:

[b * a = c * a] ) b=c

* 1 2 a 4

b 1 2 3 4

2 2 4 1 3

c 3 1 4 2

4 4 3 2 1

The row permutation property is equivalent to the left cancellation

property:

[a * b = a *c] ) b=c

* b 2 c 4

1 1 2 3 4

2 2 4 1 3

a 3 1 4 2

4 4 3 2 1

Z5* = {1,2,3,4}

*5 1 2 3 4

1 1 2 3 4

2 2 4 1 3

3 3 1 4 2

4 4 3 2 1

Euler Phi Function

(n) = size of zn*

= number of 1<k<n that are relatively prime to n.

p prime ) Zp*= {1,2,3,…,p-1}

) (p) = p-1

The additive inverse of a2 Zn is the unique b2 Zn such that a +n b ´n 0. We denote this inverse by “–

a”.

It is trivial to calculate: “-a” = (n-a).

The multiplicative inverse of a2 Zn* is

the unique b2 Zn* such that

a *n b ´n 1. We denote this inverse by “a-1” or “1/a”.

The unique inverse of a must exist because the a row contains a permutation of the

elements and hence contains a unique 1.* 1 b 3 4

1 1 2 3 4

2 2 4 1 3

a 3 1 4 2

4 4 3 2 1

Zn = {0, 1, 2, …, n-1}Zn

* = {x 2 Zn | GCD(x,n) =1}

Define +n and *n:

a +n b = (a+b mod n) a *n b = (a*b mod n)

c *n ( a +n b) ´n (c *n a) +n (c*n b)

<Zn, +n>1. Closed2. Associative3. 0 is identity4. Additive Inverses5. Cancellation6. Commutative

<Zn*, *n>

1. Closed2. Associative3. 1 is identity4. Multiplicative Inverses5. Cancellation6. Commutative

The multiplicative inverse of a2 Zn* is the unique b2

Zn* such that

a *n b ´n 1. We denote this inverse by “a-1” or “1/a”.

Efficient algorithm to compute a-1 from a and n.

Execute the Extended Euclid Algorithm on a and n (previous lecture). It will give two

integers r and s such that:ra + sn = (a,n) = 1

Taking both sides mod n, we obtain:rn ´n 1

Output r, which is the inverse of a

What is 1/29 mod 67?

Euclid’s Extended GCD algorithm

Input: X,Y Output: r,s,d such that rX+sY = d = GCD(X,Y) 67=<1,0>

29=<0,1>Euclid(67,29) 9=67 – 2*29 9 =<1,-2>Euclid(29,9) 2=29 – 3*9 2=<-3,7>Euclid(9,2) 1=9 – 4*2 1=<13,-30>Euclid(2,1) 0=2 – 2*1 2=<-3,7>

Euclid(1,0) outputs 1 = 13*67 – 30*29

HENCE 1/29 = 30.

Fundamental lemma of powers?

If (a ´n b) Then xa ´n xb ?

If (a ´n b) Then xa ´n xb ?

NO!

(16 ´15 1) , but it is not the case that: 21 ´15 216

Two names for the same set:

Zn* = Zn

a

Zna = {a *n x | x 2 Zn

*}, a2 Zn*

* b 2 c 4

1 1 2 3 4

2 2 4 1 3

a 3 1 4 2

4 4 3 2 1

Two products on the same set:

Zn* = Zn

a

Zna = {a *n x | x 2 Zn

*}, a2 Zn*

x ´n ax [as x ranges over Zn* ]

x ´n x (asize of Zn*) [Commutativity]

1 = asize of Zn* [Cancellation] a(n) = 1

Euler’s Theorem

a2 Zn*, a(n) ´n 1

Fermat’s Little Theorem

p prime, a2 Zp*) ap-1 ´p 1

(pq) = (p-1)(q-1) if p,q distinct primes

pq = # of numbers from 1 to pqp = # of multiples of q up to pqq = # of multiples of p up to pq1 = # of multiple of both p and q up to pq

(pq) = pq – p – q + 1 = (p-1)(q-1)

Fundamental lemma of powers.

Suppose x2 Zn*, and a,b,n are naturals.

If a ´(n) b Then xa ´n xb

Equivalently,xa mod (n) ´n xb mod (n)

Zn = {0, 1, 2, …, n-1}Zn

* = {x 2 Zn | GCD(x,n) =1}

Quick raising to power.

<Zn, +n>1. Closed2. Associative3. 0 is identity4. Additive Inverses

Fast + amd -5. Cancellation6. Commutative

<Zn*, *n>

1. Closed2. Associative3. 1 is identity4. Multiplicative Inverses

Fast * and /5. Cancellation6. Commutative

Euler Phi Function

(n) = size of zn*

= number of 1<k<n that are relatively prime to n.

p prime ) Zp*= {1,2,3,…,p-1}

) (p) = p-1

(pq) = (p-1)(q-1) if p,q distinct primes

pq = # of numbers from 1 to pqp = # of multiples of q up to pqq = # of multiples of p up to pq1 = # of multiple of both p and q up to pq

(pq) = pq – p – q + 1 = (p-1)(q-1)

The RSA Cryptosystem

Rivest, Shamir, and Adelman (1978)

RSA is one of the most used cryptographic protocols on the net. Your browser uses it to establish a

secure session with a site.

Pick secret, random k-bit primes: p,q “Publish”: n = p*q (n) = (p) (q) = (p-1)*(q-1)

Pick random e Z*(n)

“Publish”: eCompute d = inverse of e in Z*

(n)

Hence, e*d = 1 [ mod (n) ]

“Private Key”: d

n,e is my public key.

Use it to send a message to

me.

p,q random primes, e random Z*(n)

n = p*qe*d = 1 [ mod (n) ]

n,e

p,q prime, e random Z*(n)

n = p*qe*d = 1 [ mod (n) ]

m

n,e


n = p*qe*d = 1 [ mod (n) ]

m

me [mod n]

n,e


n = p*qe*d = 1 [ mod (n) ]

m

me [mod n]

(me)d ´n m

modular arithmetic

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