Math 546 – Congruence The relation

�

x ! ymodn means that

�

x ! y is a multiple of n (or equivalently that n divides

�

x ! y ). For any integer n >1, the relation

�

x ! ymodn is an equivalence relation on the set of all integers.

�

x ! ymodn is reflexive since for any integer m,

�

n |(m !m) and so

�

m ! mmodn .

�

x ! ymodn is symmetric since if

�

x ! ymodn then

�

x ! y = kn for some integer k and hence,

�

y ! x = !k( )n and so

�

y ! xmodn .

�

x ! ymodn is transitive since if

�

x ! ymodn and

�

y ! zmod n , then

�

x ! y = kn, y ! z = qn for some integers k and q. But then

�

x ! z = x ! y( ) + y ! z( ) = kn + qn = (k + q)n , and so

�

x ! zmod n . Properties of

�

x ! ymodn . I. Suppose that

�

x ! ymodn and k is any integer. Then (i).

�

k + x ! k + ymod n (ii).

�

kx ! kymod n i.e., The result of adding or multiplying on both sides of a congruence by the same constant is also a congruence. Make sure that you can verify this result.

II. Suppose that

�

x ! ymodn and

�

a ! bmod n . Then (i).

�

a + x ! b + ymod n (ii).

�

ax ! bymod n i.e., the result of adding or multiplying corresponding sides of a congruence is again a congruence. Proof. (i). Since

�

x ! ymodn and

�

a ! bmod n , there exist integers k and q such that

�

x ! y = kn, a ! b = qn , and so

�

a + x( ) ! b + y( ) = kn ! qn = k ! q( )n and hence

�

a + x ! b + ymod n . (ii). Since

�

x ! ymodn , then by I(ii), we get

�

ax ! aymodn . Also since

�

a ! bmod n , we get from I(ii) that

�

ay ! bymod n . But now, since congruence modulo n is transitive, it follows from

�

ax ! aymodn and

�

ay ! bymod n that

�

ax ! bymod n .

III. For every positive integers n and m there exist integers q and r such that

�

m = nq + r. 0 ! r < n . (This effectively just says that if m is divided by n, then we get some quotient q and remainder r.) So it is clear that if a and b are integers that leave the same remainder when divided by n, then

�

a ! bmod n . Conversely, if

�

a ! bmod n , then a and b leave the same remainder when divided by n.

IV. Suppose that

�

x � ymodn and k is any positive integer. Then

�

xk� y

kmodn .

Proof. This follows from II(ii) by induction. For the result is clearly true when k = 1. And if, for some k ≥ 1, it is true that

�

xk� y

kmodn , then since

�

x ! ymodn it follows that

�

n �nk� x�x

kmodn � n

k+ 1� x

k +1 .

Example So now suppose that we want to find the remainder when

�

21000 is divided by 7.

We note that

�

37� 2mod 7 and so

�

23( )333

� 1333mod 7 � 2

999� 1mod 7 .

But then

�

2 !2999

" 2 !1mod 7# 21000

" 2 mod 7 . So the remainder is 2. A Harder Example What is the remainder when

�

7100 is divided by 17? This is harder but not too bad. We

might notice first that

�

72� �2 mod 17 since

�

51 =17 � 3. But now squaring both sides of the last congruence we get,

�

74! 4 mod 17 . Now squaring both sides again we get.

�

78� 16 mod 17 and since

�

16 � �1 mod 17 we get by transitivity that

�

78� �1 mod 17 .

But now squaring both sides of this last congruence gives us

�

716� 1mod 17 .

Hence we get

�

716( )

6

� 16mod 17 � 7

96� 1mod 17 . And now since

�

796! 6mod 67

and

�

74� 4 mod 17 , we get

�

7100

! d moo 17 . So the remainder when

�

7100 is divided by

17 is 4. Equivalence Classes For the relation

�

x � ymodn , the equivalence class of an integer m is simply the set of all those integers that leave the same remainder as m when divided by n. Example: For

�

x � ymod 6 ,

�

[ 5 ] = �, � 7, � 1, 5, 11, 17, 23,�{ } ,

�

[ 2 ] = !, !10, ! 4, 2, 8, 14, 20,!{ } Polynomials If

�

P(c) = c0

+ c1c + c

2c2

+!ckck is any polynomial with integer coefficients, then if

�

a � b mod n , it follows that

�

P(a) ! P(b) mod n . This follows directly from properties I and IV above. Example. Since

�

3m� d mod33 it follows that

�

135

+ 6�133

+ 7 � 25

+ 6 �23

+ 7 mod11 . (In this case the polynomial is

�

P(x) + x5

6x3

7 .