Lecture #2. Euclid’s algorithm. Congruences.

Definition (MNZ Def. 1.2) For b, c P Z (b � 0 or c � 0),pb, cq = gcdpb, cq :� the greatest common divisor of b and c .

1

Lecture #2. Euclid’s algorithm. Congruences.

Definition (MNZ Def. 1.2) For b, c P Z (b � 0 or c � 0),pb, cq = gcdpb, cq :� the greatest common divisor of b and c .

Similarly for b1, b2, . . . , bn P Z (at least one � 0),

pb1, b2, . . . , bnq = gcdpb1, b2, . . . , bnq :� the greatest commondivisor of b1, b2, . . . , bn.

2

Lecture #2. Euclid’s algorithm. Congruences.

Definition (MNZ Def. 1.2) For b, c P Z (b � 0 or c � 0),pb, cq = gcdpb, cq :� the greatest common divisor of b and c .

Similarly for b1, b2, . . . , bn P Z (at least one � 0),

pb1, b2, . . . , bnq = gcdpb1, b2, . . . , bnq :� the greatest commondivisor of b1, b2, . . . , bn.

CAUTION: As you know, “pb1, . . . , bnq” sometimes just refersto “the n-tuple of integers b1, . . . , bn”, and not to their gcd. Itshould be clear from the context which is meant.

3

Ex: p10, 15q �

4

Ex: p10, 15q � 5 � p10,�15q � p�10,�15q.

5

Ex: p10, 15q � 5 � p10,�15q � p�10,�15q.

Ex: p1000, 700q �

6

Ex: p10, 15q � 5 � p10,�15q � p�10,�15q.

Ex: p1000, 700q � 100

7

Ex: p10, 15q � 5 � p10,�15q � p�10,�15q.

Ex: p1000, 700q � 100

Ex:

�

25 � 310 � 54, 27 � 38 � 72

�

�

8

Ex: p10, 15q � 5 � p10,�15q � p�10,�15q.

Ex: p1000, 700q � 100

Ex:

�

25 � 310 � 54, 27 � 38 � 72

�

� 25 � 38.

9

Ex: p10, 15q � 5 � p10,�15q � p�10,�15q.

Ex: p1000, 700q � 100

Ex:

�

25 � 310 � 54, 27 � 38 � 72

�

� 25 � 38.

Ex: If a �±

p pαppq and b �

±

p pβppq, then

gcdpa, bq �

10

Ex: p10, 15q � 5 � p10,�15q � p�10,�15q.

Ex: p1000, 700q � 100

Ex:

�

25 � 310 � 54, 27 � 38 � 72

�

� 25 � 38.

Ex: If a �±

p pαppq and b �

±

p pβppq, then

gcdpa, bq �±

p pminpαppq,βppqq.

11

Ex: p10, 15q � 5 � p10,�15q � p�10,�15q.

Ex: p1000, 700q � 100

Ex:

�

25 � 310 � 54, 27 � 38 � 72

�

� 25 � 38.

Ex: If a �±

p pαppq and b �

±

p pβppq, then

gcdpa, bq �±

p pminpαppq,βppqq.

More generally, if b1, . . . , bn P N, bj �±

p pβjppq, then

gcdpb1, b2, . . . , bnq �±

p pminpβ1ppq,...,βnppqq.

12

Also define:rb1, b2, . . . , bns = lcmpb1, b2, . . . , bnq :� the least common multiple

of b1, b2, . . . , bn.

Note that if bj �±

p pβjppq then

lcmpb1, b2, . . . , bnq �±

p pmaxpβ1ppq,...,βnppqq.

13

Theorem 1 (MNZ Thm. 1.3):

If g � gcdpb, cq then Dx0, y0 P Z such that g � bx0 � cy0.

14

Theorem 1 (MNZ Thm. 1.3):

If g � gcdpb, cq then Dx0, y0 P Z such that g � bx0 � cy0.

Proof. (We follow the proof in MNZ.) Set

I � tbx � cy : x, y P Zu.

15

Theorem 1 (MNZ Thm. 1.3):

If g � gcdpb, cq then Dx0, y0 P Z such that g � bx0 � cy0.

Proof. (We follow the proof in MNZ.) Set

I � tbx � cy : x, y P Zu.

Note 0 P I and I X Z� � H (since b � 0 or c � 0).

16

Theorem 1 (MNZ Thm. 1.3):

If g � gcdpb, cq then Dx0, y0 P Z such that g � bx0 � cy0.

Proof. (We follow the proof in MNZ.) Set

I � tbx � cy : x, y P Zu.

Note 0 P I and I X Z� � H (since b � 0 or c � 0).

Set

ℓ � minpI X Z�q.

17

Theorem 1 (MNZ Thm. 1.3):

If g � gcdpb, cq then Dx0, y0 P Z such that g � bx0 � cy0.

Proof. (We follow the proof in MNZ.) Set

I � tbx � cy : x, y P Zu.

Note 0 P I and I X Z� � H (since b � 0 or c � 0).

Set

ℓ � minpI X Z�q.

We will prove that ℓ � g � gcdpb, cq! (Then we are done!)

18

Theorem 1 (MNZ Thm. 1.3):

If g � gcdpb, cq then Dx0, y0 P Z such that g � bx0 � cy0.

Proof. (We follow the proof in MNZ.) Set

I � tbx � cy : x, y P Zu.

Note 0 P I and I X Z� � H (since b � 0 or c � 0).

Set

ℓ � minpI X Z�q.

We will prove that ℓ � g � gcdpb, cq! (Then we are done!)

Take x0, y0 P Z so that ℓ � bx0 � cy0.

19

Our situation: I � tbx � cy : x, y P Zu

and

ℓ � minpI X Z�q � bx0 � cy0.

Want to prove: ℓ � g � gcdpb, cq.

20

Our situation: I � tbx � cy : x, y P Zu

and

ℓ � minpI X Z�q � bx0 � cy0.

Want to prove: ℓ � g � gcdpb, cq.

We first prove ℓ | b:

21

Our situation: I � tbx � cy : x, y P Zu

and

ℓ � minpI X Z�q � bx0 � cy0.

Want to prove: ℓ � g � gcdpb, cq.

We first prove ℓ | b:

Div algo ñ Dq, r P Z such that b � ℓq � r , 0 ¤ r   ℓ.

Then r � b � ℓq � b � pbx0 � cy0qq P I. Hence r � 0, by thedefinition of ℓ.

Hence b � ℓq, i.e. we have proved that ℓ | b.

22

Our situation: I � tbx � cy : x, y P Zu

and

ℓ � minpI X Z�q � bx0 � cy0.

Want to prove: ℓ � g � gcdpb, cq.

We first prove ℓ | b:

Div algo ñ Dq, r P Z such that b � ℓq � r , 0 ¤ r   ℓ.

Then r � b � ℓq � b � pbx0 � cy0qq P I. Hence r � 0, by thedefinition of ℓ.

Hence b � ℓq, i.e. we have proved that ℓ | b.

By symmetry we also have ℓ | c .

23

Our situation: I � tbx � cy : x, y P Zu

and

ℓ � minpI X Z�q � bx0 � cy0.

Want to prove: ℓ � g � gcdpb, cq.

We have proved ℓ | b and ℓ | c .

24

Our situation: I � tbx � cy : x, y P Zu

and

ℓ � minpI X Z�q � bx0 � cy0.

Want to prove: ℓ � g � gcdpb, cq.

We have proved ℓ | b and ℓ | c .

Also g | ℓ, since ℓ � bx0 � cy0 and g | b and g | c .

Hence g ¤ ℓ.

25

Our situation: I � tbx � cy : x, y P Zu

and

ℓ � minpI X Z�q � bx0 � cy0.

Want to prove: ℓ � g � gcdpb, cq.

We have proved ℓ | b and ℓ | c . (*)

Also g | ℓ, since ℓ � bx0 � cy0 and g | b and g | c .

Hence g ¤ ℓ. (**)

Together, (*) and (**) imply g � ℓ � bx0 � cy0, QED!

26

Remark The set

I � tbx � cy : x, y P Zu

which appears in the above proof is an ideal in Z, i.e. it satisfies

(1) �s, t P I: s � t P I and (2) �r P Z, s P I: r s P I.

In fact it follows easily from the above proof that

I � gZ � tgx : x P Zu

(Cf. LL pp. 2–4 and/or KF Sec. 4.)

27

Theorem 2 (MNZ Thm. 1.4):

gcdpb, cq � min

�

tbx � cy : x, y P Zu X Z�

�

�

the positive common divisor of b and c which isdivisible by every common divisor of b and c

�

.

“Proof”: Clear/easy from Theorem 1!

28

Theorem 2 (MNZ Thm. 1.4):

gcdpb, cq � min

�

tbx � cy : x, y P Zu X Z�

�

�

the positive common divisor of b and c which isdivisible by every common divisor of b and c

�

.

Theorem 3 (MNZ Thms 1.6-10):

(a) For m P Z

�: pma,mbq � mpa, bq.

(b) If d P Z� and d | a and d | b, then

�a

d,b

d

�

1

d

pa, bq.

(b’) If d � pa, bq then

�a

d,b

d

� 1.

(c) If pa,mq � pb,mq � 1 then pab,mq � 1.(d) If c | ab and pb, cq � 1 then c | a.(e) For x P Z: pa, bq � pa, b � axq.

29

Theorem 2 (MNZ Thm. 1.4):

gcdpb, cq � min

�

tbx � cy : x, y P Zu X Z�

�

�

the positive common divisor of b and c which isdivisible by every common divisor of b and c

�

.

Theorem 3 (MNZ Thms 1.6-10):

(a) For m P Z

�: pma,mbq � mpa, bq.

(b) If d P Z� and d | a and d | b, then

�a

d,b

d

�

1

d

pa, bq.

(b’) If d � pa, bq then

�a

d,b

d

� 1.

(c) If pa,mq � pb,mq � 1 then pab,mq � 1.(d) If c | ab and pb, cq � 1 then c | a.(e) For x P Z: pa, bq � pa, b � axq.

— Part of our “toolbox”!30

Theorem 2 (MNZ Thm. 1.4):

gcdpb, cq � min

�

tbx � cy : x, y P Zu X Z�

�

�

the positive common divisor of b and c which isdivisible by every common divisor of b and c

�

.

Theorem 3 (MNZ Thms 1.6-10):

(a) For m P Z

�: pma,mbq � mpa, bq.

(b) If d P Z� and d | a and d | b, then

�a

d,b

d

�

1

d

pa, bq.

(b’) If d � pa, bq then

�a

d,b

d

� 1.

(c) If pa,mq � pb,mq � 1 then pab,mq � 1.(d) If c | ab and pb, cq � 1 then c | a.(e) For x P Z: pa, bq � pa, b � axq.

— Proof: “easy” from Theorems 1 and 2...31

Theorem 3 (MNZ Thms 1.6-10):

(a) For m P Z

�: pma,mbq � mpa, bq.

(b) If d P Z� and d | a and d | b, then

�a

d,b

d

�

1

d

pa, bq.

(b’) If d � pa, bq then

�a

d,b

d

� 1.

(c) If pa,mq � pb,mq � 1 then pab,mq � 1.(d) If c | ab and pb, cq � 1 then c | a.(e) For x P Z: pa, bq � pa, b � axq.

32

Theorem 3 (MNZ Thms 1.6-10):

(a) For m P Z

�: pma,mbq � mpa, bq.

(b) If d P Z� and d | a and d | b, then

�a

d,b

d

�

1

d

pa, bq.

(b’) If d � pa, bq then

�a

d,b

d

� 1.

(c) If pa,mq � pb,mq � 1 then pab,mq � 1.(d) If c | ab and pb, cq � 1 then c | a.(e) For x P Z: pa, bq � pa, b � axq.

Note that Theorem 3(d) implies the Key Lemma of Lecture #1:

If p, a, b P Z and p | ab and p is a prime, then p | a or p | b.

This completes the proof of the Fundamental Theorem ofArithmetic (i.e. Theorem 1.3, about unique prime factorization)!

33

Theorem 3 (MNZ Thms 1.6-10):

(a) For m P Z

�: pma,mbq � mpa, bq.

(b) If d P Z� and d | a and d | b, then

�a

d,b

d

�

1

d

pa, bq.

(b’) If d � pa, bq then

�a

d,b

d

� 1.

(c) If pa,mq � pb,mq � 1 then pab,mq � 1.(d) If c | ab and pb, cq � 1 then c | a.(e) For x P Z: pa, bq � pa, b � axq.

Note that Theorem 3(d) implies the Key Lemma of Lecture #1:

If p, a, b P Z and p | ab and p is a prime, then p | a or p | b.

This completes the proof of the Fundamental Theorem ofArithmetic (i.e. Theorem 1.3, about unique prime factorization).

On the other hand: Note that if we assume unique prime factor-ization, then Theorem 3 is “immediate”!

34

Theorem 3 (MNZ Thms 1.6-10):

(a) For m P Z

�: pma,mbq � mpa, bq.

(b) If d P Z� and d | a and d | b, then

�a

d,b

d

�

1

d

pa, bq.

(b’) If d � pa, bq then

�a

d,b

d

� 1.

(c) If pa,mq � pb,mq � 1 then pab,mq � 1.(d) If c | ab and pb, cq � 1 then c | a.(e) For x P Z: pa, bq � pa, b � axq.

Definition (MNZ Def 1.3): We say that a and b are relativelyprime if pa, bq � 1.

Similarly for an n-tuple of integers b1, . . . , bn, we say that b1, . . . , bnare relatively prime if pb1, . . . , bnq � 1. A stronger property is tosay that b1, . . . , bn are pairwise relatively prime; this means that

pbi , bjq � 1 for all 1 ¤ i   j ¤ n.

35

Next, how to compute gcdpb, cq?

36

Next, how to compute gcdpb, cq?

— Theorem 3(e), pa, bq � pa, b � axq , is a good tool!

37

Next, how to compute gcdpb, cq?

— Theorem 3(e), pa, bq � pa, b � axq , is a good tool!

Ex:

p1105, 117q �

38

Next, how to compute gcdpb, cq?

— Theorem 3(e), pa, bq � pa, b � axq , is a good tool!

Ex:

p1105, 117q � p1105� 9 � 117, 117q � p52, 117q �

(We used 1105117 � 9.4....)

39

Next, how to compute gcdpb, cq?

— Theorem 3(e), pa, bq � pa, b � axq , is a good tool!

Ex:

p1105, 117q � p1105� 9 � 117, 117q � p52, 117q �

� p52, 117� 2 � 52q � p52, 13q �.

40

Next, how to compute gcdpb, cq?

— Theorem 3(e), pa, bq � pa, b � axq , is a good tool!

Ex:

p1105, 117q � p1105� 9 � 117, 117q � p52, 117q �

� p52, 117� 2 � 52q � p52, 13q � 13.

41

Next, how to compute gcdpb, cq?

— Theorem 3(e), pa, bq � pa, b � axq , is a good tool!

Ex:

p1105, 117q � p1105� 9 � 117, 117q � p52, 117q �

� p52, 117� 2 � 52q � p52, 13q � 13.

— Note that by following the above computation backwards, onecan also compute integers x, y such that 1105x � 117y � 13.

42

Theorem 4 (Euclid’s Algorithm):

Given b, c P Z�, use the division algorithm to obtain

b � cq1 � r1 0   r1   c

c � r1q2 � r2 0   r2   r1r1 � r2q2 � r3 0   r3   r2

� � � � � �

rj�2 � rj�1qj � rj 0   rj   rj�1rj�1 � rjqj�1.

Then gcdpb, cq � rj .

(Special case: If “j � 0”, i.e. r1 � 0, then gcdpb, cq � c .)

x0, y0 P Z giving gcdpb, cq � bx0 � cy0 can be obtained bysuccessively expressing r1, r2, . . . , rj as linear combinations of b, c .

43

Theorem 4 (Euclid’s Algorithm):

Given b, c P Z�, use the division algorithm to obtain

b � cq1 � r1 0   r1   c

c � r1q2 � r2 0   r2   r1r1 � r2q2 � r3 0   r3   r2

� � � � � �

rj�2 � rj�1qj � rj 0   rj   rj�1rj�1 � rjqj�1.

Then gcdpb, cq � rj .

(Special case: If “j � 0”, i.e. r1 � 0, then gcdpb, cq � c .)

x0, y0 P Z giving gcdpb, cq � bx0 � cy0 can be obtained bysuccessively expressing r1, r2, . . . , rj as linear combinations of b, c .

Proof: Clear using pa, bq � pa, b � axq (=Theorem 3(e)).The process ends since r1 ¡ r2 ¡ � � � ¡ 0.

44

Ex: gcdp1105, 117q � 13, as we computed above via

1105 � 117 � 9� 52

117 � 52 � 2� 13

52 � 13 � 4.

45

Ex: gcdp1105, 117q � 13, as we computed above via

1105 � 117 � 9� 52

117 � 52 � 2� 13

52 � 13 � 4.

Hence gcdp1105, 117q � 13 � 117� 2 � 52

46

Ex: gcdp1105, 117q � 13, as we computed above via

1105 � 117 � 9� 52

117 � 52 � 2� 13

52 � 13 � 4.

Hence gcdp1105, 117q � 13 � 117� 2 � 52

� 117� 2p1105� 9 � 117q �

47

Ex: gcdp1105, 117q � 13, as we computed above via

1105 � 117 � 9� 52

117 � 52 � 2� 13

52 � 13 � 4.

Hence gcdp1105, 117q � 13 � 117� 2 � 52

� 117� 2p1105� 9 � 117q � p�2q � 1105� 19 � 117.

Thus: we have 13 � 1105x0 � 117y0 for x0 � �2, y0 � 19.

(Other examples: MNZ pp. 12–15, LL p. 6.)

48

Ex: gcdp1105, 117q � 13, as we computed above via

1105 � 117 � 9� 52

117 � 52 � 2� 13

52 � 13 � 4.

Hence gcdp1105, 117q � 13 � 117� 2 � 52

� 117� 2p1105� 9 � 117q � p�2q � 1105� 19 � 117.

Thus: we have 13 � 1105x0 � 117y0 for x0 � �2, y0 � 19.

(Other examples: MNZ pp. 12–15, LL p. 6.)

Remark: j   3 log c in Theorem 4 (cf. MNZ p. 15).

49

Congruences.

Definition: Let m P Z

� and a, b P Z.

We say a � b pmod mq if m | a � b.

50

Congruences.

Definition: Let m P Z

� and a, b P Z.

We say a � b pmod mq if m | a � b.

Same thing:

am

� b,

a � bpmq,

“a is congruent to b mod m”.

51

Theorem 5 (MNZ Thm. 2.1 + more!):

Let m P Z

� and a, b, c, d P Z. Then:

(1) am

� b � bm

� a � a � bm

� 0.

(2)

�

am

� b and bm

� c

�

ñ am

� c.

(3) am

� a

(4) the relationm

� is an equivalence relation on Z.

(5)

�

am

� b and cm

� d�

ñ a � cm

� b � d.

(6)

�

am

� b and cm

� d�

ñ acm

� bd.

(7)

�

a � b pmod mq and d | m, d ¡ 0

�

ñ a � b pmod dq.(8)

�

a � b pmod mq and c ¡ 0�

ñ ac � bc pmod mcq.

52

Definition: For m P Z

� and a P Z we write

a :�

b P Z : bm

� a

(

= the residue class of a.

53

Definition: For m P Z

� and a P Z we write

a :�

b P Z : bm

� a

(

= the residue class of a.

(Note that m is implicit in this notation; KF calls the same thing“Rmpaq”.)

54

Definition: For m P Z

� and a P Z we write

a :�

b P Z : bm

� a

(

= the residue class of a.

Also:

Zm :� the set of residue classes pmod mq.

55

Definition: For m P Z

� and a P Z we write

a :�

b P Z : bm

� a

(

= the residue class of a.

Also:

Zm :� the set of residue classes pmod mq.

Ex: Modulo 3 we have

0 �

. . . ,�6,�3, 0, 3, 6, . . .

(

1 �

. . . ,�5,�2, 1, 4, 7, . . .

(

2 �

. . . ,�4,�1, 2, 5, 8, . . .

(

,

and

Z3 �

0, 1, 2(

.

56

Definition: For m P Z

� and a P Z we write

a :�

b P Z : bm

� a

(

= the residue class of a.

Also:

Zm :� the set of residue classes pmod mq.

Ex/fact: For a general m P Z

�, we have

Zm �

0, 1, 2, . . . ,m � 1

(

�

1, 2, 3, . . . ,m

(

,

and here for each 0 ¤ r   m we have that r equals the set ofall integers which give remainder r when divided by m (using theDivision Algorithm, Thm. 1.2).

57

Definition The operations � and � on residue classes: For fixedm P Z

�, we define the operations � and � in Zm by:

a � b :� a � b p�a, b P Zq;

a � b :� a � b p�a, b P Zq;

and also�a :� �a and a � b :� a � b.

58

Definition The operations � and � on residue classes: For fixedm P Z

�, we define the operations � and � in Zm by:

a � b :� a � b p�a, b P Zq;

a � b :� a � b p�a, b P Zq;

and also�a :� �a and a � b :� a � b.

These operations are well-defined since, by Theorem 5(5),(6):

�

am

� b and cm

� d

�

ñ a � cm

� b � d , and

�

am

� b and cm

� d

�

ñ acm

� bd.

59

Definition The operations � and � on residue classes: For fixedm P Z

�, we define the operations � and � in Zm by:

a � b :� a � b p�a, b P Zq;

a � b :� a � b p�a, b P Zq;

and also�a :� �a and a � b :� a � b.

These operations are well-defined since, by Theorem 5(5),(6):

�

am

� b and cm

� d

�

ñ a � cm

� b � d , and

�

am

� b and cm

� d

�

ñ acm

� bd.

Ex: In Z8, 1� 3 � 4 � 9� 27 � 36, which is ok!

60

Definition The operations � and � on residue classes: For fixedm P Z

�, we define the operations � and � in Zm by:

a � b :� a � b p�a, b P Zq;

a � b :� a � b p�a, b P Zq;

and also�a :� �a and a � b :� a � b.

These operations are well-defined since, by Theorem 5(5),(6):

�

am

� b and cm

� d

�

ñ a � cm

� b � d , and

�

am

� b and cm

� d

�

ñ acm

� bd.

Ex: In Z8, 1� 3 � 4 � 9� 27 � 36, which is ok!

Remark (extracurricular): xZm,�, �y is a ring, cf. KF §3 and §5.

61

A large part of the course is about computing in Zm, solvingequations in Zm, and interesting facts and formulas in Zm whichdon’t have a counterpart in Z or R!

62

Ex: For a, b, x,m P Z with m ¡ 0, we have

a � x � b pmod mq �

63

Ex: For a, b, x,m P Z with m ¡ 0, we have

a � x � b pmod mq � a � x � am

� b � a� xm

� b � a

Thus we have solved the equation a � x � b pmod mq !

64

Ex: ax � 1 pmod mq?

65

Ex: ax � 1 pmod mq?

For m � 3;

the multiplication table of Z3:

� 0 1 2

0 0 0 0

1 0 1 2

2 0 2 1

66

Ex: ax � 1 pmod mq?

For m � 3;

the multiplication table of Z3:

� 0 1 2

0 0 0 0

1 0 1 2

2 0 2 1

ñ

�

ax � 1 pmod 3q�

has a solution x iff a � 1 or 2 pmod 3q.

67

Ex: ax � 1 pmod mq?

For m � 4;

the multiplication table of Z4:

� 0 1 2 3

0 0 0 0 0

1 0 1 2 3

2 0 2 0 2

3 0 3 2 1

68

Ex: ax � 1 pmod mq?

For m � 4;

the multiplication table of Z4:

� 0 1 2 3

0 0 0 0 0

1 0 1 2 3

2 0 2 0 2

3 0 3 2 1

ñ

�

ax � 1 pmod 4q�

has a solution x iff a � 1 or 3 pmod 4q.

69

Ex: ax � 1 pmod mq?

For m � 4;

the multiplication table of Z4:

� 0 1 2 3

0 0 0 0 0

1 0 1 2 3

2 0 2 0 2

3 0 3 2 1

ñ

�

ax � 1 pmod 4q�

has a solution x iff a � 1 or 3 pmod 4q.

Theorem 6 (MNZ Thm 2.9): Let a P Z, m P Z

�. The equationax � 1 pmod mq has a solution x pmod mq iff pa,mq � 1. Whenthis holds, the solution is unique mod m.

70

Theorem 6 (MNZ Thm 2.9): Let a P Z, m P Z

�. The equationax � 1 pmod mq has a solution x pmod mq iff pa,mq � 1. Whenthis holds, the solution is unique mod m.

Def: Z�m � ta : a P Z, gcdpa,mq � 1u.

71

Theorem 6 (MNZ Thm 2.9): Let a P Z, m P Z

�. The equationax � 1 pmod mq has a solution x pmod mq iff pa,mq � 1. Whenthis holds, the solution is unique mod m.

Def: Z�m � ta : a P Z, gcdpa,mq � 1u.

(Here note that pa,mq only depends on a P Zm; namely if a, a

1

are any two integers with a � a1 then pa,mq � pa1,mq.)

72

Theorem 6 (MNZ Thm 2.9): Let a P Z, m P Z

�. The equationax � 1 pmod mq has a solution x pmod mq iff pa,mq � 1. Whenthis holds, the solution is unique mod m.

Def: Z�m � ta : a P Z, gcdpa,mq � 1u.

(Here note that pa,mq only depends on a P Zm; namely if a, a

1

are any two integers with a � a1 then pa,mq � pa1,mq.)

Def: We say that α P Zm is invertible if α P Z

�

m. For α P Z

�

m

we write α�1 for the unique solution to αx � 1.

(Note α�1 P Z�m.)

(Cf. LL Def. 4.11 “relatively prime”, and KF Sec. 6.)

73

Theorem 6 (MNZ Thm 2.9): Let a P Z, m P Z

�. The equationax � 1 pmod mq has a solution x pmod mq iff pa,mq � 1. Whenthis holds, the solution is unique mod m.

Def: Z�m � ta : a P Z, gcdpa,mq � 1u.

(Here note that pa,mq only depends on a P Zm; namely if a, a

1

are any two integers with a � a1 then pa,mq � pa1,mq.)

Def: We say that α P Zm is invertible if α P Z

�

m. For α P Z

�

m

we write α�1 for the unique solution to αx � 1.

(Note α�1 P Z�m.)

Note: α, β P Z�m ñ

�

αβ P Z�m and pαβq�1 � α�1β�1

�

.

Hence Z�m is a group. (Cf. MNZ 2.10-11; this is extracurricular.)

74

Theorem 6 (MNZ Thm 2.9): Let a P Z, m P Z

�. The equationax � 1 pmod mq has a solution x pmod mq iff pa,mq � 1. Whenthis holds, the solution is unique mod m.

Proof:

75

Theorem 6 (MNZ Thm 2.9): Let a P Z, m P Z

�. The equationax � 1 pmod mq has a solution x pmod mq iff pa,mq � 1. Whenthis holds, the solution is unique mod m.

Proof:

If pa,mq � 1 then Dx, y P Z such that ax �my � 1;

76

Theorem 6 (MNZ Thm 2.9): Let a P Z, m P Z

�. The equationax � 1 pmod mq has a solution x pmod mq iff pa,mq � 1. Whenthis holds, the solution is unique mod m.

Proof:

If pa,mq � 1 then Dx, y P Z such that ax �my � 1;

thus ax � 1 pmod mq.

77

Theorem 6 (MNZ Thm 2.9): Let a P Z, m P Z

�. The equationax � 1 pmod mq has a solution x pmod mq iff pa,mq � 1. Whenthis holds, the solution is unique mod m.

Proof:

If pa,mq � 1 then Dx, y P Z such that ax �my � 1;

thus ax � 1 pmod mq.

Conversely, if ax � 1 pmod mq then Dy P Z such that

ax �my � 1; thus pa,mq � 1 by Theorem 2.

78

Theorem 6 (MNZ Thm 2.9): Let a P Z, m P Z

�. The equationax � 1 pmod mq has a solution x pmod mq iff pa,mq � 1. Whenthis holds, the solution is unique mod m.

Proof:

If pa,mq � 1 then Dx, y P Z such that ax �my � 1;

thus ax � 1 pmod mq.

Conversely, if ax � 1 pmod mq then Dy P Z such that

ax �my � 1; thus pa,mq � 1 by Theorem 2.

Finally, if ax1 � ax2 � 1 pmod mq, then

79

Theorem 6 (MNZ Thm 2.9): Let a P Z, m P Z

�. The equationax � 1 pmod mq has a solution x pmod mq iff pa,mq � 1. Whenthis holds, the solution is unique mod m.

Proof:

If pa,mq � 1 then Dx, y P Z such that ax �my � 1;

thus ax � 1 pmod mq.

Conversely, if ax � 1 pmod mq then Dy P Z such that

ax �my � 1; thus pa,mq � 1 by Theorem 2.

Finally, if ax1 � ax2 � 1 pmod mq, then

x1 a

loomoon

�1

x1 � x1 a

loomoon

�1

x2 pmod mq,

and so

x1 � x2 pmod mq.

Done! �

80

Ex: Solve the congruence equation 103x � 1 pmod 143q.

81

Ex: Solve the congruence equation 103x � 1 pmod 143q.

Solution: Is p103, 143q � 1? Compute using Euclid’s algorithm!

82

Ex: Solve the congruence equation 103x � 1 pmod 143q.

Solution: Is p103, 143q � 1? Compute using Euclid’s algorithm!

143 � 103 � 1� 40

103 � 40 � 2� 23

40 � 23 � 1� 17

23 � 17 � 1� 6

17 � 6 � 2� 5

6 � 5 � 1� 1

Hence YES, p103, 143q � 1.

83

Ex: Solve the congruence equation 103x � 1 pmod 143q.

Solution: Is p103, 143q � 1? Compute using Euclid’s algorithm!

143 � 103 � 1� 40

103 � 40 � 2� 23

40 � 23 � 1� 17

23 � 17 � 1� 6

17 � 6 � 2� 5

6 � 5 � 1� 1

Hence YES, p103, 143q � 1.

Now use the above to find x, y P Z with 103x � 143y � 1:

84

143 � 103 � 1� 40

103 � 40 � 2� 23

40 � 23 � 1� 17

23 � 17 � 1� 6

17 � 6 � 2� 5

6 � 5 � 1� 1

Hence YES, p103, 143q � 1.

Now use the above to find x, y P Z with 103x � 143y � 1:

1 � 6� 5

� 6� p17� 2 � 6q � p�1q � 17� 3 � 6

� p�1q � 17� 3 � p23� 17q � 3 � 23� 4 � 17

� 3 � 23� 4 � p40� 23q � �4 � 40� 7 � 23

� �4 � 40� 7 � p103� 2 � 40q � �18 � 40� 7 � 103

� 7 � 103� 18 � p143� 103q � �18 � 143� 25 � 103.

85

Ex: Solve the congruence equation 103x � 1 pmod 143q.

Solution: .........

We have proved that p103, 143q � 1, and found that

103x � 143y � 1 holds with x � �18 and y � 25.

86

Ex: Solve the congruence equation 103x � 1 pmod 143q.

Solution: .........

We have proved that p103, 143q � 1, and found that

103x � 143y � 1 holds with x � �18 and y � 25.

Hence 103 � p�18q � 1 pmod 143q,

87

Ex: Solve the congruence equation 103x � 1 pmod 143q.

Solution: .........

We have proved that p103, 143q � 1, and found that

103x � 143y � 1 holds with x � �18 and y � 25.

Hence 103 � p�18q � 1 pmod 143q, and by Theorem 6,

x � �18 pmod 143q is the unique solution to the equation

�

103x � 1 pmod 143q�

.

Answer: The unique solution is x � �18 pmod 143q

(or equivalently, x � 125 pmod 143q).

88

Finally, a result about how to “divide both sides with a” whensolving a congruence equation:

89

Finally, a result about how to “divide both sides with a” whensolving a congruence equation:

Theorem 7 (MNZ Thm. 2.3, LL Prop. 4.5):

Let a, x, y P Z, m P Z

�. Then

ax � ay pmod mq � x � y

�

modm

pm, aq

.

In particular, if pa,mq � 1, then

ax � ay pmod mq � x � y pmod mq.

90

Finally, a result about how to “divide both sides with a” whensolving a congruence equation:

Theorem 7 (MNZ Thm. 2.3, LL Prop. 4.5):

Let a, x, y P Z, m P Z

�. Then

ax � ay pmod mq � x � y

�

modm

pm, aq

.

In particular, if pa,mq � 1, then

ax � ay pmod mq � x � y pmod mq.

Ex: 21x � 21y pmod 35q � x � y pmod 5q.

15x � 15y pmod 10q � x � y pmod 2q.

91

Theorem 7

Let a, x, y P Z, m P Z

�. Then

ax � ay pmod mq � x � y

�

modm

pm, aq

.

In particular, if pa,mq � 1, then

ax � ay pmod mq � x � y pmod mq.

Proof:

92

Theorem 7

Let a, x, y P Z, m P Z

�. Then

ax � ay pmod mq � x � y

�

modm

pm, aq

.

In particular, if pa,mq � 1, then

ax � ay pmod mq � x � y pmod mq.

Proof: First note that the second statement is indeed a specialcase of the first statement.

(One may also note that the second statement can be proveddirectly – using Theorem 6 – by “multiplying the equation witha�1”.)

93

Theorem 7

Let a, x, y P Z, m P Z

�. Then

ax � ay pmod mq � x � y

�

modm

pm, aq

.

.......................

Proof: We now prove the first statement in Theorem 7.

First assume ax � ay pmod mq.

Then ax � ay � mz for some z P Z.

94

Theorem 7

Let a, x, y P Z, m P Z

�. Then

ax � ay pmod mq � x � y

�

modm

pm, aq

.

.......................

Proof: We now prove the first statement in Theorem 7.

First assume ax � ay pmod mq.

Then ax � ay � mz for some z P Z.

Hencea

pa,mqpx � y q �

m

pa,mqz , and so

m

pa,mq�

��

a

pa,mqpx � y q.

95

Theorem 7

Let a, x, y P Z, m P Z

�. Then

ax � ay pmod mq � x � y

�

modm

pm, aq

.

.......................

Proof: We now prove the first statement in Theorem 7.

First assume ax � ay pmod mq.

Then ax � ay � mz for some z P Z.

Hencea

pa,mqpx � y q �

m

pa,mqz , and so

m

pa,mq�

��

a

pa,mqpx � y q.

Also

� m

pa,mq

,a

pa,mq

�

pm, aq

pa,mq� 1 pThm. 3(b)q,

96

Theorem 7

Let a, x, y P Z, m P Z

�. Then

ax � ay pmod mq � x � y

�

modm

pm, aq

.

.......................

Proof: We now prove the first statement in Theorem 7.

First assume ax � ay pmod mq.

Then ax � ay � mz for some z P Z.

Hencea

pa,mqpx � y q �

m

pa,mqz , and so

m

pa,mq�

��

a

pa,mqpx � y q.

Also

� m

pa,mq

,a

pa,mq

�

pm, aq

pa,mq� 1 pThm. 3(b)q,

and som

pa,mq�

��

x � y , i.e. x � y

�

modm

pa,mq

.

97

Theorem 7

Let a, x, y P Z, m P Z

�. Then

ax � ay pmod mq � x � y

�

modm

pm, aq

.

.......................

Proof: Conversely, assume x � y

�

modm

pa,mq

.

Then by Thm. 5(8):

ax � ay�

modam

pa,mq

,

and heream

pa,mq� m �

a

pa,mq

is divisible by m;

hence by Theorem 5(7) we get:

ax � ay pmod mq.

�

98