Download Modular Arithmetic Basics (1) The “floor” function is defined by the

Modular Arithmetic Basics
(1) The “floor” function is defined by the formula
bxc := ( the greatest integer less than or equal to x).
This is also known as “the greatest integer function,” and in old texts is denoted by
(whole) brackets. Examples: b3.789c = 3; b−3.789c = −4.
(2) The “greatest common divisor,” abbreviated gcd, of a set of integers is, of course, the
largest positive integer that divides every integer in the set. Examples: gcd(24, 52) =
4; gcd(54, 42) = 6.
(3) The “mod” operator is defined as follows:
x mod m := x − m · bx/mc
if m 6= 0. For positive integers x and m, x mod m = the remainder in integer division
of x by m. Examples: 110 mod 26 = 6; −52 mod 26 = 0.
(4) The “mod” relation is defined as follows:
a ≡ b (mod m) if and only if a mod m = b mod m.
The above definitions make sense even for real numbers. When a, b, m are integers
and m > 0,
a ≡ b (mod m) if and only if a − b is a multiple of m.
Examples: 110 ≡ 6 (mod 26); −80 ≡ 24 (mod 26).
(5) Modular arithmetic and algebra behave “as expected” for the operations of addition,
subtraction, and multiplication. If a ≡ b (mod m) and c ≡ d (mod m), then a + c ≡
b + d (mod m), a − c ≡ b − d (mod m), and ac ≡ bd (mod m).
(6) But division and cancellation are trickier. Here are the cancellation rules.
ad ≡ bd (mod m) if and only if a ≡ b (mod m), assuming gcd(d, m) = 1.
ad ≡ bd (mod m0 d) if and only if a ≡ b (mod m0 ), assuming d 6= 0.
Combined: ad ≡ bd (mod m) if and only if a ≡ b (mod m/ gcd(d, m)).
(7) Let a, x, m be integers with m > 0. Let g = gcd(a, m). The number of solutions of
ax ≡ b (mod m) in the set {1, 2, . . . , m} is 0 if g does not divide b and is g if g|b,
and then if x0 is one solution, then all solutions are given by x = x0 + (m/g)k for
k = . . . , −2, −1, 0, 1, 2, 3, . . ., and g of them are in a complete set of residues.
(8) Let a, b be integers. Then a|b, read “a divides b,” if and only if b is a multiple,
i.e., an integer multiple, of a: b = ka for some integer k. Examples: 7|98; −5|100;
but 8 does not divide 26. Graham, Knuth, and Patashnik’s divisibility notation:
a\b if and only if a > 0 and a|b.
(9) A positive integer p is called prime if it has just two divisors, namely 1 and p. The
unending sequence of primes starts thus: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, . . .. The
Fundamental Theorem of Arithmetic: There is only one way to write a positive
integer as a product of primes in nondecreasing order. Examples: 12 = 2 · 2 · 3,
715 = 5 · 11 · 13.
1
(10) Notation: Write m ⊥ n if gcd(m, n) = 1. This is read as “m is prime to n,” and we
say m and n are “relatively prime.” Of course, if m ⊥ n, then n ⊥ m too.
(11) The Euler phi function (or totient function) is defined for positive integers n by
φ(n) := #{k|0 < k ≤ n, gcd(k, n) = 1}, i.e., φ(n) is the number of positive integers
less than or equal to n that are prime to n.
n
1 2 3 4 5 6 7 8 9 10 11 12
φ(n) 1 1 2 2 4 2 6 4 6 4 10 4
If p and q are distinct primes, then φ(p) = p − 1 and φ(pq) = (p − 1)(q − 1).
(12) Euler’s Theorem: If a ⊥ n, then aφ(n) ≡ 1 (mod n).
The special case when n = p, a prime, is known as
Fermat’s Little Theorem: If a ⊥ p, then ap−1 ≡ 1 (mod p).
(13) The Chinese Remainder Theorem (Sun Tsŭ, c. A.D. 350):
If m ⊥ n, then a ≡ b (mod mn) ⇐⇒ a ≡ b (mod m) and a ≡ b (mod n).
(14) The extended Euclidean algorithm finds the greatest common divisor g of positive
integers a and b (with a > b) and integers s and t such that g = sa + tb. The
calculation may be arranged in a tableau. At the start one has
q r s t
a 1 0
q b 0 1
where q = ba/bc, the integer quotient of a divided by b. Subsequently row j is
calculated as follows.
rj = rj−2 − qj−1 · rj−1 ,
sj = sj−2 − qj−1 · sj−1 ,
tj = tj−2 − qj−1 · tj−1 ,
qj = brj−1 /rj c.
The process ends when the remainder rj = 0.
For example, gcd(7469, 2387) = 77 = 8 × 7469 − 25 × 2387:
q
r
s
t
7469
1
0
3 2387
0
1
7 308
1 −3
1 231 −7
22
3
77
8 −25
0
The extended Euclidean algorithm may be used to solve ax ≡ b (mod n) when
g := gcd(a, n)|b: Calculate g = sa + tn. Then x ≡ sb/g + (n/g)k (mod n) with
k = 0, ±1, ±2, . . .. When b = 1, this procedure finds reciprocals mod n.