SECTION 1DIVISION ALGORTHM. Given positive intergers a and b, b 0 and such that a = bq+r and 0≤r<b. EUCLIDEAN ALGORITHM. If a and b are positive intergers and b 0 and we know that a = bq+r b=rq1+r1 r=r1q2+r2 and and and . . rk=rk+1qk+2+rk+2 and 0≤rk+2<rk+1 rt-1=rtqt+1 and (a,b)=rt 0≤r<b 0≤r1<r 0≤r2<r1 For sufficiently large k, say k=t, It follows that the sequence of numbers b>r>r1>r2>……eventually goes to 0. It also follows that there exists a linear combination of (a,b). THEOREM 4. If (a,b)=d, ax+by=d. PROOF: Go backwards using the Euclidean Algorithm. COROLLARY 1. If and only if d|ab and (d,a)=1 then d|b. intergers x,y such that Exercises 1 1. Calculate (314,159) and (4144, 7696) a. 314=159+155 i. 159=155+4 ii. 155=38*4+1 iii. 4=3+1 iv. GCD=1 b. 7696=4144+3552 i. 4144=3552+592 ii. 3552=592*6 iii. GCD=592 2. Calculate (3141, 1592) and (10001,100083) a. 3141=1592+1549 i. 1592=1549+43 ii. 1549=43(36)+1 iii. 43=1*43 iv. GCD=1 b. 100083=10001*10+73 abc). we see that f | N and f | c. c).b)=(N. we see that d | N and d | a. from our assumption. Direct Proof: i. Then e | (N . b) = (N. and f = (N.y Z. we see that e | abc. (N. 592=3552-592*5 b. GCD=73 3. y=1. Since f | c. Then we know. 13=299*5-247*6 i. a). 6. d | 1. Case 1: x=-1. d | N and d | abc. Since b|b1=x(y) ii. Thus. we see that d | 1. Thus. Proof by contradiction: i. 592=[4144-7696+4144] e. we see that d | abc. y=-40 4. x=79. y=-1 s. c) = 1. Then f | (N . Assume N=abc+1. 10001=73*137 d. a. Proceeding by contradiction. Find two different solutions of 299x+247y=13 a. 1=39*(159-155)-155 e. Since f = (N. b).abc). Find x and y such that 4144x+7696y=592 a. Proof : i. We must prove (N. Since d = (N. 1=79*159-40*159 h. Since a=bx b=(bx)y. 1=39*159-40*155 f. e | 1. s.c)=d such that ii.y)=(5. It follows that a=bx and b=ay.a)=(N. Then d = 1. a contradiction. c). 13=(299-247)-(247-(299-247*4)) d.x. a).t. xy=1 . If N=abc+1(N.a)=(N.c)=1. Thus. 13=(52)-(247-52*4) c. 13=(299-247*2+299*4-247*4 e. xy=1 1. 1=4-1(155-38*4) c.t. Since e | b. Then f = 1. Case 2: x=1. Assume a|b and b|a and that a.b)=(N. f | 1. 1=39*4-155 d. Prove that if a|b and b|a a=b or a=-b.c)=1 a. Since e = (N. (x. Then d | (N .c. Then e = 1. d≠1 iii. we see that f | abc. suppose (N. a) = (N. It follows that a=b(-1)=-b and further that b=a(-1) = -a iii. Thus. Since d | a. e = (N. 592=[4144-3552](1) d.b)=(N. Find x and y such that 314x+159y=1 a.a)=(N. Let d = (N.-6) 7. b. 592=[592(6-5)] c. 13=52-39 b. Thus. we see that e | N and e | b.abc). 1=39*159-40*(314-159) g. 1=4-1*3 b. b).b. 592=2*4144-1*7696 5. Note: 1 is neither prime nor composite but do refer to it as a unit. If n2=1. Proof: From Lemma 1. n>1 can be written as aproduct of primes.t. say d. . That isn1=p2n2 where p2 is a prime and 1≤n2<n1. then we are done: n=p1 is an expression of n as aproduct of primes. It follows with this and the statement a|b that (a. then n is prime by definition. we will have nk=1 in which case n=p1p2p3…. then we are done.b)=a. b. Case 1: The set is empty. Prove that if a|b and a>0 then (a. then so would n. If x2+ax+b=0 has an rational root show that it is an integer. If d had a disor greter than 1 and less than d. That is n=p1n1 where 1≤n1<n. c<m. Definition: An interger that is gretater than 1 but is not prime is called a composite precisely because it is a combination of primes. Case 1: gcd(a. c. For some k.b)=a. a. 15. It is either empty or nonempty. a. n>1 is divisible by a prime. Note that the same prime may occur several times in the product. 8. b. iv.1. is positive. (a. Thus a|b and b|a implies that a=b or a=-b. then Lemma 1 once again states that n2=p3n3 where p3 is a prime and 1≤n3<n2. such a sequence cannot continue forever with intergers. If not we continue. Lemma 3: If n is compositie. a.m)=1 implies that a and m have no common factors. Thus di is prime and n has a prime divisor. Proof: Consider the set of divisors of n that are less than n but greater than 1. Lemma 2: Every interger n.pk is the desired expression of n as a product of primes. If n3=1.t. If x2+ax+b=0 has an interger root show that it divides b. If n1>1. we know that there is aprime p1 s. While c|b it still stands that since c<a. Case 2: gcd(a. Lemma 1: Every interger n.b)=a. Assume a|b and a>0 and that a. It follows that the set of intergers can be divided into three classes: primes. composites. then it has a divisor d such that 1<d≤n1/2. namely d. If n1=1. It follows that a=b and further that b=a. Section 2: Unique Factorization Definition: A prime is an interger that is greaterthan 1 and has no positive divisors other than itself and 1. c<a. Case 2: The set is nonempty. p1|n.m)=c≠1 implies that cx=a and cy=m s. then from Lemma 1 again there is a prime that divides n1. There are infinitely many primes.m Z. then the least-interger principle states that it has a smallest element. thus n has a prime divsor—itself.b. We will sooner or later come to one of the ni=1 because n>n1>n2>… and each n. and a unit. It follows that b=am. Theorem 1 (Euclid). but this is impossible because d was the sallest such divisor. Lemma 4: If n is composite then it has a prime divisor less than or equal to n1/2... where a finite number of the ni are positive integers. and p|q1q2…..2. Theorem 2: The Unique Factorization Theorem: Any positive interger can be written as a product of primes in one and only one way. i=1.. 4. 1 is represented by the empty product.. For example 999 = 33×37. or the standard form of n. 1000 = 23×53. 1001 = 7×11×13 Note that factors p0 = 1 may be inserted without changing the value of n (e.qnthen p=qk forsome k. This is the basis for a proof by induction.k.pn and b = q1q2. Allowing negative exponents provides a canonical form for positive rational numbers. If n is prime. . 1000 = 23×30×53).qn are primes... where n = ab and 1 < a ≤ b < n. then p|a or p|b... there are integers a and b.. a = p1p2. then p|ai for some i. 3.3…. .. But then n = ab = p1p2. Proof: The proof uses Euclid's lemma (Elements VII. This representation is called the canonical representation of n. The article has proofs of the lemma. q2. Assume it is true for all numbers less than n.... Canonical representation of a positive integer Every positive integer n > 1 can be represented in exactly one way as a product of prime powers: where p1 < p2 < .qm is the product of primes. By the induction hypothesis. Lemma 7: If q1.pnq1q2. Uniqueness Assume that s > 1 is the product of prime numbers in two different ways: We must show m = n and that the qj are a rearrangement of the pi.. [edit] Existence By inspection.…. 30): if a prime p divides the product of two natural numbers a and b.. 2. Lemma 6: If p|a1a2. any positive integer can be uniquely represented as an infinite product taken over all the positive prime numbers. is the product of primes.ak. then p divides a or p divides b (or perhaps both). each of the small natural numbers 1. Lemma 5: If p|ab. < pk are primes and the αi are positive integers. In fact.qm are products of primes.. there is nothing more to prove. Otherwise. and the rest are zero.g. A linear Diophantine equation is an equation between two sums of monomials of degree zero or one. Lemma 1. they define an algebraic curve. or more general object. a Diophantine equation is an indeterminate polynomial equation that allows the variables to take integer values only. Therefore m = n and every qj is a pi.y0 is a solution of ax+by=c. who made a study of such equations and was one of the first mathematicians to introduce symbolism into algebra. Definition: Prime power decomposition is where numbers are broken down in terms of their prime factors and numbers are written as products of prime factors including the use of index form. Section 3: Linear Diophantine Equations In mathematics. and ask about the lattice points on it. showing that m ≤ n. p2 must equal one of the remaining qj. relabeling the qj if necessary. Relabeling again if necessary. The word Diophantine refers to the Hellenistic mathematician of the 3rd century. say that p1 divides q1. p1 = q1. Diophantus of Alexandria. While individual equations present a kind of puzzle and have been considered throughout history. so that Reasoning the same way. since the product of numbers greater than 1 cannot equal 1.b)|c. say p2 = q2. Proof: We are given that ax0+by0=c. If there were any qj left over we would have which is impossible. Thus . In more technical language. algebraic surface. The mathematical study of Diophantine problems Diophantus initiated is now called "Diophantine analysis". But q1 is prime. If x0.By Euclid's lemma p1 must divide one of the qj. Diophantine problems have fewer equations than unknown variables and involve finding integers that work correctly for all equations. Therefore. The equation ax+by=c has an interger solution if and only if (a. so its only divisors are itself and 1. the formulation of general theories of Diophantine equations (beyond the theory of quadratic forms) was an achievement of the twentieth century. then so is x0+bt. Then This can be done for all m of the pi. y0-at forany interger t. y=2-2t and x=[y-t]/2 3. y. For larger integer values of n. the other letters being given are constants.z): the Pythagorean triples. and z.a(x0+bt)+b(y0-at)=ax0+abt+by0-bat =ax0+by0 =c So x0+bt. Although not usually stated in polynomial form. Find all integer solutions a. y. . for every positive integer n ≥ 2. this example is equivalent to the polynomial equation 4xyz = yzn + xzn + xyn = n(yz + xz + xy). Yes 17(61)=1037 c. x+y=2 x=2-t and y=2-t b.y.99 4. It follows that this is a Diophantine equation and is in fact solvable. a. Does (51. all as positive integers. 2x+y=2. 85)=17 b. Does 17|1037? i. therefore.7. 3|5+4t when t=…1. =51*2-85(-1) d. y. (Pell's equation) which is named after the English mathematician John Pell. For n = 2 there are infinitely many solutions (x.7. z). y0-at satisfies the equation too. p+10d+25q=4. =51-[85-51]*1 ii. Find 5 ways to use 100 coins to get 4. This is a linear Diophantine equation (see the section "Linear Diophantine equations" below).85) exist? i. The Erdős–Straus conjecture states that. Examples of Diophantine equations In the following Diophantine equations.11…4n-1 2.…. Fermat's Last Theorem states there are no positive integer solutions (x. 3x-4y=5 x=[5+4t]/3 and y=[3t-5]/4 i. as well as by Fermat in the 17th century. (51. Exercises 3 1. dimes. =51r+85s f.3n-2 ii. 1037=51(122)+85(-61) e. It was studied by Brahmagupta in the 7th century. 4|3t-5 when t=…3. Find all itnerger solutions a. x. there exists a solution in x. and z are the unknowns.99 with pennies. 51x+85y=1037 a. 17=51-34*1 i.4. and quarters. a b(mod m) there exists an interger k such that a=b+km. Proof: Suppose that a b(mod m). Then.m|(a-b). From the definition of divisibility . Every integer is congruent (mod m) . Theorem 1. we know that since there is an interger k such that km=a-b then a=b+km. form the definition of congruence. suppose that a=b+km. Conversely.i. SECTION 4: CONGRUENCES We say that a is congruent to b modulo m (a b(mod m)) m|(a-b). Theorem 2.