Number Theory and Graph Theory Chapter 2 Prime numbers and PDF

Definition 1.Letnbe a positive integer anda;b2Z. Thenaandbare said to becongruent modulo norais said to becongruent tobmodulon, denotedab(modn), ifndividesab. That is, there exists k2Zsuch that ab=kn: Since 1 divides every integer. So any two integers are congruent modulo 1. Two integers are congruent modulo 2 if and only if either both are even or both are odd. Leta2Z;n2N:Then, by division algorithm we havea=nq+r, where0r1be a fixed positive integer and let a;b;c2Z. Then, the following hold:

1. aa(modm).

2. If ab(modm), then ba(modm).

3. If ab(modm)and bc(modm), then ac(modm).

4. If ab(modm)and cd(modm), then acbd(modm)and acbd(modm).

5. If ab(modm), then a+cb+c(modm)and acbc(modm)for any c2Z.

6. If ab(modm), then anbn(modm)for any positive integer n.

7. If acbc(modm), then ab(modmd

), where d=gcd(c;m).

8. If ab(modm), and kjm then ab(modk).

9. If ab(modm)and c2N, then cacb(modcm).

10. If ab(modm)and the integers a;b;m are all divisible by d>0, thenad

bd (modmd

Proof.Proof of Part 1: Sincemj0=aa,aa(modm).Proof of Part 2: Sinceab(modm), somjab. Hence,ab=mq. Thus,mjm(q)=ba.

Hence,ba(modm).

Proof of Part 3: Ifab(modm)andbc(modm)thenmjabandmjbc. Hence, by the linearity propertymjac= (ab)+(bc)and thusac(modm). Proof of Part 4: Sinceac=a+(c), it suffices to prove only the "+case." By assumption, mjabandmjcd. Therefore, by linearity,mj(a+c)(b+d) = (ab)+(cd)and m jc(ab)+b(cd) =acbd. Hence a+cb+d(modm)andacbd(modm): ProofofPart5: Sinceab(modm),mjab. Thus,mjc(ab)andmj(a+ccb)=ab.

Hence,a+cb+c(modm)andacbc(modm).

4 Proof of Part 6: We proveanbn(modm)by induction onn.

Ifn=1, the result is true by the assumption thatab(modm).Assume that the result holds forn=k. That is,akbk(modm). We also haveab(modm).

Thus,aakbbk(modm)or equivalently,ak+1bk+1(modm). Hence, by the Principle of Mathematical Induction (PMI), the result holds for alln2N. Proof of Part 7: Asacbc(modm), we getmjacbd=c(ab). Thus,c(ab) =mk, for somek2Z. Since,(c;m) =d,c=dk1andm=dk2, for somek1;k22Z. Thus,dk1(ab) =dk2 ork2jabas gcd(k1;k2) =1:

Thus,ab(modk2=md

Proof of Part 8, 9 and 10 are left for the readers.Definition 3. Letm2Nbe a given. For eacha2Z, theresidue class(or thecongruence classor equivalence class ) of amodulom, denoted[a]or[a]m, is defined as [a] =fx2Zjxa(modm)g: Thus, the setf[0];[1];[2];:::;[m1]g, denotedZm, has some nice properties.

Theorem 4.

Letp(x) =må

k=0ckxkbe a polynomial function ofxwith integral coefficientsck. Ifab (modn), then p(a)p(b) (modn):

Proof.

Since,ab(modn), we have seen thatakbk(modn)and hence,ckakckbk(modn), fork=0;1;2;:::;m. Adding thesem+1 congruences, we get p (a) =må k=0c kakmå k=0c kbk=p(b) (modn): Ifp(x)is a polynomial with integral coefficients, we say thatais a solution of the congruence p (x)0(modn)ifp(a)0(modn):

Corollary 5.

Ifais a solution ofp(x)0(modn)andab(modn), thenbis also a solution of p (x)0(modn): 5 Theorem6.LetM=am10m+am110m1++10a1+a0be the decimal expansion of the positive integer M,0ak<10, and let S=a0+a1++am:Then,9jM if and only if9jS.

Proof.

Letp(x) =må

k=0akxk. Thenp(10) =Mandp(1) =S. But,101(mod 9)and hence p (10)p(1) (mod 9). Thus, we haveMS(mod 9):Theorem7. LetM=am10m+am110m1++10a1+a0be the decimal expansion of the positive integer M,0ak<10, and let T=a0a1++(1)mam:Then,11jM if and only if11jT.

Proof.

Letp(x) =må

k=0akxk. Thenp(10) =Mandp(1) =T. As10 1(mod 11), we get p (10)p(1) (mod 11)and hence,MT(mod 11):Fermat"s Little Theorem

It is easy to see that

41(mod 5); 241(mod 5); 341(mod 5); 441(mod 5)

40(mod 5)

41(mod 5); 741(mod 5); 841(mod 5); 941(mod 5)

40(mod 5)

Theorem 8.

[Fermat"s Little Theorem] Letpbe a prime and suppose thatp-a. Thenap11 (modp):

Proof.

We begin by considering the firstp1positive multiples ofa. That is, consider the integers a ;2a;3a;:::;(p1)a: None of these numbers is congruent to another modulop. Letrasa(modp)for1rOr equivalently, a p

1(p1)!(p1)!(modp):

Since, gcd

(p;(p1)!) =1, using Theorem 2.thm:procon:7, we haveap11(modp):Corollary 9.If p is prime, then apa(modp)for any integer a.

Proof.Ifpja, thenpjapaand hence the result is true.Ifp-a, then using theorem 8,ap11(modp):Now, multiplying both sides bya, we getapa

(modp).Alternate proof: The result is clearly true forp=2as bothaanda2have the same parity. Let pbe an odd prime, thenapandahave same sign. Thus, it is sufficient to prove the result for positive integers. So, let us fix a primepand prove the result using induction ona:Ifa=1, then clearly apa(modp)holds. Assume the result holds fora,i.e.,apa(modp). We need to prove that(a+1)p(a+1) (modp).

We first observe that sincepis a primepjp

k=p!k!(pk)!fork=1;2;:::;p1. Hence, (a+1)pap+1(modp). But, by induction hypothesis,apa(modp). Hence, we get (a+1)pap+1(a+1) (modp).quotesdbs_dbs14.pdfusesText_20

[PDF] a crash course in c

[PDF] a d s solutions pvt ltd bangalore

[PDF] a d s solutions pvt ltd zauba

[PDF] a dialogue between a teacher and a student about studies

[PDF] a feasible solution for

[PDF] a feasible solution to an lp problem

[PDF] a feasible solution to linear programming problem should

[PDF] a final class can be abstract

[PDF] a final class can be extended

[PDF] a final class can be extended. true false

[PDF] a final class can have subclass i.s. it can be extended

[PDF] a final method can be inherited

[PDF] a first course in graph theory pdf

[PDF] a fois b au carré

[PDF] a for apple to z for

[PDF] Number Theory and Graph Theory Chapter 2 Prime numbers and