1. Applied Symbolic Computation (CS 300) Modular Arithmetic
September 4, 1997
Applied Symbolic Computation (CS 300) Modular Arithmetic
Jeremy R. Johnson

2. September 4, 1997
Introduction
Objective: To become familiar with modular arithmetic and some key algorithmic constructions that are important for computer algebra algorithms.
Modular Arithmetic
Modular inverses and the extended Euclidean algorithm
Fermat’s theorem
Euler’s Identity
Chinese Remainder Theorem

3. Modular Arithmetic (Zn)
September 4, 1997
Modular Arithmetic (Zn)
Definition: a  b (mod n)  n | (b - a)
Alternatively, a = qn + b
Properties (equivalence relation)
a  a (mod n) [Reflexive]
a  b (mod n)  b  a (mod n) [Symmetric]
a  b (mod n) and b  c (mod n)  a  c (mod n) [Transitive]
Definition: An equivalence class mod n
[a] = { x: x  a (mod n)} = { a + qn | q  Z}

4. Example n = 5 [0] = … -10, -5, 0, 5, 10 … [1] = … -9, -4, 1, 6, 11 …
September 4, 1997
Example n = 5
[0] = … -10, -5, 0, 5, 10 …
[1] = … -9, -4, 1, 6, 11 …
[2] = … -8, -3, 2, 7, 12 …
[3] = … -7, -2, 3, 8, 13 …
[4] = … -6, -1, 4, 9, 14 …
Equivalence classes partition the set Z

5. Modular Arithmetic (Zn)
September 4, 1997
Modular Arithmetic (Zn)
It is possible to perform arithmetic with equivalence classes mod n.
[a] + [b] = [a+b]
[a] * [b] = [a*b]
In order for this to make sense, you must get the same answer (equivalence) class independent of the choice of a and b. In other words, if you replace a and b by numbers equivalent to a or b mod n you end up with the sum/product being in the same equivalence class.
a1  a2 (mod n) and b1  b2 (mod n)  a1+ b1  a2 + b2 (mod n)
a1* b1  a2 * b2 (mod n)
(a + q1n) + (b + q2n) = a + b + (q1 + q2)n
(a + q1n) * (b + q2n) = a * b + (b*q1 + a*q2 + q1* q2)n

6. September 4, 1997
Representation of Zn
The equivalence classes [a] mod n, are typically represented by the representatives a.
Positive Representation: Choose the smallest positive integer in the class [a] then the representation is {0,1,…,n-1}.
Symmetric Representation: Choose the integer with the smallest absolute value in the class [a]. The representation is {-(n-1)/2 ,…, n/2 }. When n is even, choose the positive representative with absolute value n/2.
E.G. Z6 = {-2,-1,0,1,2,3}, Z5 = {-2,-1,0,1,2}

7. Example n = 5 Note multiplicative inverses. i.e. ab  1 (mod n)
September 4, 1997
Example n = 5
0 |
1 |
2 |
3 |
4 | *
0 |
1 |
2 |
3 |
4 |
Note multiplicative inverses. i.e. ab  1 (mod n)

8. Example n = 6 Note zero divisors (product of non-zero elts equal to 0)
September 4, 1997
Example n = 6
0 |
1 |
2 |
3 |
4 |
5 | *
0 |
1 |
2 |
3 |
4 |
5 |
Note zero divisors (product of non-zero elts equal to 0)

9. The Euclidean Algorithm
September 4, 1997
The Euclidean Algorithm
Definition. Integer division with remainder
a = qb+r 0  r < b
b|a if a = qb
Definition: Greatest Common Divisor
g = gcd(a,b).
g|a and g|b.
If e|a and e|b then e|g
The Euclidean Algorithm provides and efficient method for computing gcd(a,b)
gcd(a,0) = a
gcd(a,b) = gcd(b, a mod b)

10. Bezout’s Identity Definition. Bezout’s Identity Proof.
September 4, 1997
Bezout’s Identity
Definition. Bezout’s Identity
There exist integers x, y such that ax+by = gcd(a,b)
Proof.
Let  = {ax+by, x,y  Z} and let d  Z have smallest abs value.
Claim d = gcd(a,b).
Note that a,b   a+b and s Z, a   sa 
a = qd + r, 0  r < d and r = a – qd    r = 0 and d|a. Similarly d|b.
If e|a and e|b then e|(ax+by)  e|d.

11. Remainder Sequence Definition. Remainder Sequence
September 4, 1997
Remainder Sequence
Definition. Remainder Sequence
a1 = a, a2 = b, a3,…, an+1 with
a  b > 0, a1 = a, a2 = b
a1 = q1a2 + a3, 0  a3 < a2 …
ai = qiai+1 + ai+2, 0  ai+2 < ai+1
an-1 = qn-1an + an+1, 0  an+1 < an
an = qnan+1, an+1
an+1 = gcd(a,b)

12. The Extended Euclidean Algorithm
September 4, 1997
The Extended Euclidean Algorithm
Let a1 = a, a2 = b, a3,…, an+1 be a remainder sequence defined by
ai = qiai+1 + ai+2, 0  ai+2 < ai+1 for i=3,…,with an+2 = 0
Definition. Cosequences
x1 = 1, x2 = 0, xi+2 = xi - qi xi+1
y1 = 0, y2 = 1, yi+2 = yi - qi yi+1
Then axi + byi = ai and in particular axn+1 + byn+1 = an+1 = gcd(a,b)
The proof is by induction.

13. Computing Time of Euclidean Alg
September 4, 1997
Computing Time of Euclidean Alg
ai = qiai+1 + ai+2, 0  ai+2 < ai+1
ai/ai+2 = qiai+1/ai  ai/ai+2 > qi + 1  2
an/an+1  2
a1/a3  a2/a4  a3/a5  …  an-1/an+1  an/an+1= a1a2/(an+1an+1)
a1/a3  a2/a4  a3/a5  …  an-1/an+1  2n
 2n ≤ a1a2/(an+1an+1) ≤ (a/gcd(a,b))2 ≤ a2
 n ≤ 2lg(a)

14. September 4, 1997
Modular Inverses
Definition: x is the inverse of a mod n, if ax  1 (mod n)
The equation ax  1 (mod n) has a solution iff gcd(a,n) = 1.
By the Extended Euclidean Algorithm, there exist x and y such that ax + ny = gcd(a,n). When gcd(a,n) = 1, we get ax + ny = 1. Taking this equation mod n, we see that ax  1 (mod n). Conversely if a has an inverse mod n equal to x then ax = 1 + qn  ax + (-qn) = 1 which implies the gcd(a,n) =1.
By taking the equation mod n, we mean applying the mod n homomorphism: m Z  Zm, which maps the integer a to the equivalence class [a]. This mapping preserves sums and products.
I.E. m(a+b) = m(a) + m(b), m(a*b) = m(a) * m(b)

15. September 4, 1997
Fermat’s Theorem
Theorem: If a  0  Zp, then ap-1  1 (mod p). More generally, if
a  Zp, then ap  a (mod p).
Proof: Assume that a  0  Zp. Then
a * 2a * … (p-1)a = (p-1)! * ap-1
Also, since a*i  a*j (mod p)  i  j (mod p), the numbers a, 2a, …, (p-1)a are distinct elements of Zp. Therefore they are equal to 1,2,…,(p-1) and their product is equal to
(p-1)! mod p. This implies that
(p-1)! * ap-1  (p-1)! (mod p)  ap-1  1 (mod p).

16. Fast Powering Power(a,e,n) Computing time O(log e)
September 4, 1997
Fast Powering
Power(a,e,n)
if e = 0 then
return 1
else
b := Power(a,e div 2,n)
return b^2 * a^(e mod 2) (mod n)
Computing time O(log e)
Assuming arithmetic in Zn is O(1)

17. September 4, 1997
Euler phi function
Definition: phi(n) = #{a: 0 < a < n and gcd(a,n) = 1}
Properties:
(p) = p-1, for prime p.
(p^e) = (p-1)*p^(e-1)
 (m*n) =  (m)* (n) for gcd(m,n) = 1.
(p*q) = (p-1)*(q-1)
Examples:
(15) = (3)* (5) = 2*4 = 8. = #{1,2,4,7,8,11,13,14}
(9) = (3-1)*3^(2-1) = 2*3 = 6 = #{1,2,4,5,7,8}

18. September 4, 1997
Euler’s Identity
The number of elements in Zn that have multiplicative inverses is equal to phi(n).
Theorem: Let (Zn)* be the elements of Zn with inverses (called units). If a  (Zn)*, then a(n)  1 (mod n).
Proof. The same proof presented for Fermat’s theorem can be used to prove this theorem.

19. Chinese Remainder Theorem
September 4, 1997
Chinese Remainder Theorem
Theorem: If gcd(m,n) = 1, then given a and b there exist an integer solution to the system:
x  a (mod m) and x = b (mod n).
Proof:
Consider the map x  (x mod m, x mod n).
This map is a 1-1 map from Zmn to Zm  Zn, since if x and y map to the same pair, then x  y (mod m) and x  y (mod n). Since gcd(m,n) = 1, this implies that x  y (mod mn).
Since there are mn elements in both Zmn and Zm  Zn, the map is also onto. This means that for every pair (a,b) we can find the desired x.

20. Alternative Interpretation of CRT
September 4, 1997
Alternative Interpretation of CRT
Let Zm  Zn denote the set of pairs (a,b) where a  Zm and b  Zn. We can perform arithmetic on Zm  Zn by performing componentwise modular arithmetic.
(a,b) + (c,d) = (a+b,c+d)
(a,b)*(c,d) = (a*c,b*d)
Theorem: Zmn  Zm  Zn. I.E. There is a 1-1 mapping from Zmn onto Zm  Zn that preserves arithmetic.
(a*c mod m, b*d mod n) = (a mod m, b mod n)*(c mod m, d mod n)
(a+c mod m, b+d mod n) = (a mod m, b mod n)+(c mod m, d mod n)
The CRT implies that the map is onto. I.E. for every pair (a,b) there is an integer x such that (x mod m, x mod n) = (a,b).

21. Constructive Chinese Remainder Theorem
September 4, 1997
Constructive Chinese Remainder Theorem
Theorem: If gcd(m,n) = 1, then there exist em and en (orthogonal idempotents)
em  1 (mod m)
em  0 (mod n)
en  0 (mod m)
en  1 (mod n)
It follows that a*em + b* en  a (mod m) and  b (mod n).
Proof.
Since gcd(m,n) = 1, by the Extended Euclidean Algorithm, there exist x and y with m*x + n*y = 1. Set em = n*y and en = m*x