Module 5: Basic Number Theory Theme 1: Division b b=a ½6=4 =4 Given two integers, say a and , the quotient may or may not be an integer (e.g., but =5 = ¾:4 ½¾ ). Number theory concerns the former case, and discovers criteria upon which one can decide about divisibility of two integers. 6=¼ a b k More formally, for a we say that divides if there is another integer such that b = ka; jb and we write a . In short: ajb 9 b = ka: ¾Z if and only if k This simple definition leads to many properties of divisibility. For example, let us establish the following lemma. jb ajc aj´b · cµ Lemma 1 If a and ,then . Proof. We give a direct proof. From the definition of divisibility and the hypotheses we know that × there are integers Ø and such that b = Øa; c = ×a: Hence b · c = a´× · ص: · Ø aj´b · cµ Since × is an integer, we prove that . Exercise 5A: Prove the following two facts: jb ajbc c 1. If a ,then for all integers . jb bjc ajc 2. If a and ,then . We already noted that an integer may be or not divisible by another integer. However, when dividing one number by another there is always a quotient and a remainder. More precisely, if a and Õ Ö d are positive integers then there is a unique and such that a = dÕ · Ö Ö<d d ¼; ½;::: ;d ½ where ¼ is a remainder. Observe that the remainder can take only values . 1 Theme 2: Primes Primes numbers occupy very prominent role in number theory. A prime number Ô is an integer ½ greater than ½ that is divisible only by and itself. A number that is not prime is called composite. Example 1: The primes less than ½¼¼ are: ¾; ¿; 5; 7; ½½; ½¿; ½7; ½9 ; ¾¿; ¾9; ¿½ ; ¿7 ; 4½; 4¿ ; 47 ; 5¿; 59; 6½ ; 67; 7½; 7¿ ; 79; 8¿; 89 ; 97: How many primes are there? We first prove that there are infinite number of primes. Theorem 1. There are infinite number of primes. Proof. We provide a proof by contradiction. Actually, it is due to Euclid and it is more than 2000 ¾; ¿; 5;::: ;Ô Ô k years old. Let us assume that there is a finite number of primes, say, k where is the largest prime (there is the largest prime since we assumed there are only finitely many of them). Construct another number Å =¾¡ ¿ ¡ 5 ¡¡¡ Ô ·½ k ¾; ¿; 5;::: ;Ô which is a product of all primes plus one. First, observe that none of the primes k Å ½ can divide Å , since the remainder of dividing by any of the primes is equal to . Since every ½ number, including Å , is divisible by at least two numbers, and itself, there must be another prime, Å ¾; ¿; 5;::: ;Ô possible itself, that is not among the primes k . This contradicts the assumption that ¾; ¿; 5;::: ;Ô k are the only primes. Ò But how many primes are there smaller than Ò,where is a fixed number. This is a very difficult ÐÓg ´Òµ problem that was solved only in the last century. Basically, there are approximately about Ò= ¾5 ½¼¼ ½¼¼= ÐÓg ´½¼¼µ ¾¾ primes smaller than Ò. For example, there are primes smaller than ,and . Primes are important since every integer can be represented as a product of primes. This is known as the Fundamental Theorem of Arithmetics and we will prove it below. Example 2: Observe that ¾ ¾ ½¼¼ = ¾ ¡ ¾ ¡ 5 ¡ 5=¾ 5 ; ¿8½ = ¿ ¡ ½¾7; ¿ 888 = ¾ ¡ ¿ ¡ ¿7: Theorem 2.[Fundamental Theorem of Arithmetics ] Every positive integer can be written uniquely as the product of primes where the prime factors are written in order of increasing size, that is, if Ò is Ô <Ô < ¡¡¡ <Ô ¾ Ñ a natural numbers and ½ are distinct primes, then e e e ½ ¾ Ñ Ò = Ô ¡ Ô ¡¡¡ Ô Ñ ½ ¾ 2 e Ô Ô Ò i i where i are exponents of (i.e., the number of times occurs in the factorization of ). Proof. We give an indirect proof. Let us assume that there are two different prime factorizations of Ò, say e e e ½ ¾ Ñ Ò = Ô ¡ Ô ¡¡¡ Ô Ñ ½ ¾ d d d ½ ¾ Ö Ò = Õ ¡ Õ ¡¡¡ Õ Ö ½ ¾ Õ < ¡¡¡ <Õ Ò Ö where ½ are primes. Since we factorize the same number we must have e e d d e d ½ ¾ ½ ¾ Ñ Ö Ô ¡ Ô ¡¡¡ Ô = Õ ¡ Õ ¡¡¡ Õ : Ñ Ö ½ ¾ ½ ¾ Ô = Õ Ô 6= Õ Ô Õ ;::: ;Õ ½ ½ ½ ½ ½ Ö We first prove that ½ .If ,then can not divide any of the primes (we say Ô Õ ;::: ;Õ Ô Õ ;::: ;Õ ½ Ö ½ ½ Ö that ½ is relatively prime to all ). Indeed, since and are primes, none of them d d d ¾ ½ Ö ¡¡¡ Õ ¡ Õ Ô Ò = Õ equal, then they must be relatively prime. But, then ½ cannot divide which is Ö ¾ ½ e e e ½ ¾ Ñ Ò = Ô Ô = Õ ¡ Ô ¡¡¡ Ô ½ nonsense since . Thus, we must conclude that ½ . Ñ ½ ¾ e = d Ô = Õ ½ ½ ½ Now we prove that ½ provided that we just established above. Again, assume e ½ e <d d = e · h h>¼ Ô ½ ½ ½ contrary that ½ , say , . Then after dividing everything by we obtain ½ d e d h e ¾ ¾ Ö Ñ : ¡ Õ ¡¡¡ Õ = Õ Ô ¡¡¡ Ô Ö ½ Ñ ¾ ¾ Õ But then the right-hand side of the above is divisible by ½ while the left-hand side is not, which is impossible since there is an equality sign between the left-hand side and the right-hand side of the above. This completes the proof. How to find out whether an integer is a prime or not? Unfortunately, there is no fast way of doing it (i.e., there is no efficient algorithm), but one can use some properties of primes and composite numbers to speed up the process. Here is one useful result. Ô Ò Ò Lemma 2.If Ò is a composite integer, then has a prime divisor less than or equal to . d ½ <d <Ò Ò = d ¡ Ö Proof.SinceÒ is a composite integer, it must have a factor such that ,thatis, Ô Ô ½ d> Ò Ö> Ò where Ö> is an integer. Let us now assume contrary that and .Butthen Ô Ô d ¡ Ö> Ò Ò = Ò = dÖ Ò which is the desired contradiction since we assumed that Ò . We must conclude that has at Ô least one divisor not exceeding Ò. This divisor is prime or not. If it is not prime, it must have a Ô prime divisor, which certainly must be smaller than Ò. We can use this lemma, in its contrapositive form, to decide whether Ò is a prime or not. Indeed. Ô Ò Ò the above lemma is equivalent to: if Ò has no prime divisor less than or equal to ,then is a prime number. 3 ½¼7 Example 3: Let us show that ½¼7 is a prime number. If would be composite, then it has had prime Ô ½¼:¿4 ½¼ ¾; ¿; 5 7 ½¼7 divisor smaller than ½¼7 . Primes smaller than are ,and . None of it divides , thus it ½¼7 must be a prime number. There were several attempts to find a systematic way of computing prime numbers. Euclid sug- k ·½µ gested that ´ -st prime can be computed recursively as follows: e = ¾; ½ e = e e ¡¡¡ e ·½: ½ ¾ k ·½ k For example, the first few numbers are e = ¾·½ = ¿; ¾ e = ¾ ¡ ¿·½ = 7; ¿ e = ¾ ¡ ¿ ¡ 7·½=4¿: 4 This is an example of a recurrence that we already encountered in the previous module. All numbers computed so far are primes. But, unfortunately, e =¾¡ ¿ ¡ 7 ¡ 4¿ · ½ = ½8¼7 = ½¿ ¡ ½¿9 5 is not a prime. Ô ½ In the seventeenth century, a French mathematician Marin Marsenne suggested that ¾ is prime provided Ô is prime. Unfortunately, ½½ ¾ ½ = ¾¼47 = ¾¿ ¡ 89: From now on we shall work under the assumption that there is no easy, simple and fast algorithm to compute prime numbers. Theme 3: Greatest Common Divisor Ò Ò Ñ The largest divisor that divides both Ñ and is called the greatest common divisor of and .It Ñ; Òµ is denoted as gcd´ . Formally: Ñ; Òµ:=ÑaÜfk : k jÑ k jÒg: gcd ´ and ¿6 Example 4: What is the greatest common divisor of ¾4 and . One way of finding it is to list all ¿6 divisors of ¾4 and and pick up the largest common to both lists. For example, = f½; ¾; ¿; 4; 6; 8; ½¾; ¾4g ; divisors of ¾4 = f½; ¾; ¿; 4; 6; 9; ½¾; ½8; ¿6 g: divisors of ¿6 4 ; ¿6µ = ½¾ Thus gcd´¾4 . Another, more systematic way is to do prime factorization of both numbers and pick up the largest common factors. In our case, ¿ ¾4 = ¾ ¡ ¿; ¾ ¾ ¿6 = ¾ ¡ ¿ : Thus ¾ gcd ´¾4; ¿6µ = ¾ ¡ ¿=½¾: Generalizing the last example, let a a a ½ ¾ k Ñ = Ô Ô ¡¡¡ Ô ; ½ ¾ k b b b ½ ¾ k Ò = Ô Ô ¡¡¡ Ô ½ ¾ k be prime factorizations with possible zero exponents. Then ÑiÒfa ;b g ÑiÒfa ;b g ÑiÒfa ;b g ½ ½ ¾ ¾ k k gcd ´Ñ; Òµ=Ô Ô ¡¡¡ Ô ½ ¾ k fÜ; Ý g Ü Ý where ÑiÒ is the minimum of and . Indeed, take the last example to see that ÑiÒf¾;¿g ÑiÒf½;¾g gcd ´¾4; ¿6µ = ¾ ¿ : Ò Exercise 5B: Let us define the least common multiple of Ñ and as the smallest positive integer Ò ÐcÑ´Ñ; Òµ ÐcÑ´5; 8µ =4¼ that is divisible by both Ñ and . It is denoted as (e.g., ). Prove that for Ò any positive integers Ñ and Ñ ¡ Ò = gcd´Ñ; Òµ ¡ ÐcÑ´Ñ; Òµ: Ò We need some more definitions. Two integers, say Ñ and , may be composite but the only Ñ Ò common divisor of both is ½. In such a case we say that and are relatively prime. More generally: a ;a ;::: ;a ½ ¾ Definition 1. The integers k are pairwise relatively prime if gcd´a ;a µ=½; 8 ½ i<j k: i j Unlike finding primes, there is an efficient algorithm (a procedure) that finds the greatest common divisor.