信息安全数学2

School of Mathematics and Applied Statistics

INFO412

Mathematics and Cryptography Section2:Congruence and Residues

Definition2.1

Let m∈N.

(i)If m|(x−a),then we say that a is congru-ent to x modulo m.We write

x≡a(mod m).

(ii)If a is congruent to x modulo m,then a is called a residue of x modulo m.If,in addition, 0≤a≤m−1,then a is called the least residue of x modulo m.

(iii)Every number x has a unique least residue modulo m,which will be one of the numbers {0,1,2,...,m−1}.(Why?)Further,each of these numbers is the least residue for itself.That is,{0,1,2,...,m−1}is a complete set of residues modulo m.

Example2.2

(i)≡(mod7)and≡−(mod7).

(ii)and−are both residues of12 modulo7,and is the least residue.

(iii)For m=5,is a complete

set of residues,and any number x will be con-gruent to exactly one of these.For example,if x=111then x≡(mod5).

Remark2.3

(i)For a given m,the complete set of least residues is{0,1,2,···,m−1},but there are many other complete sets.

(ii)In general:

A is a complete set of residues

⇐⇒

there exist integers k1,k2,...,k m such that

A={k1m,k2m+1,k3m+2,...,k m m+(m−1)}.

Example2.4

For m=5,some complete sets of residues are {10,11,12,13,14},{0,11,2,13,14}

and

.

For m=7:

{14,−6,72,10,−10,−65,6}

and

.

Theorem2.5(Basic Congruence Properties) (i)a≡b(mod m)=⇒b≡a(mod m).

(ii)a≡b(mod m)and b≡c(mod m)=⇒a≡c(mod m).

(iii)a≡b(mod m)and c≡d(mod m)=⇒

a+c≡b+d(mod m),

a−c≡b−d(mod m),

a·c≡b·d(mod m),

a i≡

b i(mod m),

for any integer i≥0.

(iv)Suppose thatφis a polynomial in variables x1,x2,...,x k and thatφhas integer coefficients. Suppose also that

a1≡b1,a2≡b2,...,a k≡b k(mod m).

Then

φ(a1,a2,...,a k)≡φ(b1,b2,...,b k)(mod m).

Example2.6

Let k=2andφ(x,y)=2x2+3xy+7y+2.

Take m=5.Then3≡8(mod5)and2≡−8 (mod5).

Therefore,

2·32+3·3·2+7·2+2

≡2·82+3·8·(−8)+7·(−8)+2(mod5). Check?

Note that a≡b(mod m)and c≡d(mod m) does not in general imply a/c≡b/d(mod m) (even assuming that a/c and b/d are whole num-bers).

Example2.7

4≡10(mod6)and2≡2(mod6),but2≡5 (mod6).

Similarly,cancellation is not valid in general: ka≡kb(mod m)does not in general imply a≡b (mod m).

Example2.8

2×2≡2×4(mod4),but2≡4(mod4).

Solving Linear Congruences

Given k,l and m,we want,if possible,to solve the following congruence for x:

kx≡l(mod m).

Note2.9

If x solves kx≡l(mod m)and x≡y(mod m), then y also solves ky≡l(mod m).For this rea-son,we usually only list those solutions which are least residues.

Example2.10

Solve2x≡1(mod4)and2x≡4(mod6).