CHAPTER 6. Prime Numbers. Definition and Fundamental Results


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1 CHAPTER 6 Prime Numbers Part VI of PJE. Definition and Fundamental Results 6.1. Definition. (PJE definition ) An integer p is prime if p > 1 and the only positive divisors of p are 1 and p. If n > 1 and n is not prime, n is said to be composite. The integer 1 is neither prime nor composite Remark. The definition implies that n is composite if, and only if, n = ab for some integers a, b with 1 < a, b < n. (Proof If n = ab where a, b > 1, then n > 1 and n is not prime since a n, a 1, a n; thus, n is composite. Vice versa, if n is composite, then n > 1 and n has a positive divisor a other than 1 and n; this means that 1 < a < n and that n = ab with 1 < b = n/a < n.) Example. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19,... Aside It is a hard problem to prove that a given large integer (say with 150 digits) is prime. But some very large primes are known, such as 2 74,207,281 1 with more than 22 million decimal digits, found on 7 January 2016 by the Great Internet Mersenne Prime Search. At the time of its discovery, this was the largest known prime number. End of aside Question. What is the connection with the earlier definition of coprime? 6.3. Lemma. If p is prime and p a then p and a are coprime. Proof. PJE p.279 The only positive divisors of p are 1 and p. Out of these, p is not a divisor of a by assumption, so the only common positive divisor of p and a is 1. Hence gcd(p, a) = 1. 51
2 Prime Numbers Proposition. (PJE proposition ) Every integer n > 1 is a product of primes. (With the convention that a product can be of just one prime!) Proof PJE p.278. We prove this by strong induction. Base case: n = 2 which is a product of one prime. Indictive step: Assume that for all n such that 2 n k, n is a product of primes. Our goal is to prove that k + 1 is a product of primes. Indeed, if k + 1 is a prime, it is a product of one prime. Otherwise, k + 1 is not a prime, hence k + 1 is composite and k + 1 = ab with 1 < a, b < k + 1. But then 2 a, b k so by Inductive Hypothesis, a, b are products of primes. But then k + 1 = ab is also a product of primes. Conclusion: by strong induction, every integer n 2 is a product of primes Theorem. Euclid s property of a prime If p is a prime, then for all integers a, b, if p divides ab then p a or p b. Proof (PJE theorem but we do not assume a, b to be positive) Let a, b Z. It is enough to assume that p a and then to show that that p b. By Lemma 6.3, p is coprime to a; but p ab, so by Corollary 2.22 p b, as required. Aside This result motivates the definition of a prime in more advanced work. End of aside Example. The number 6 is not prime, because 6 has positive divisors 2, 3 which are neither 1 nor 6. Another way to prove that 6 is not prime would be to show that 6 does not possess Euclid s property of a prime. For example, 6 24, yet 24 = 3 8 and 6 3 and 6 8; hence by Theorem 6.5, 6 is not prime Corollary. If a prime p divides the product a 1 a 2... a n of integers a 1,..., a n, then p a i for some 1 i n. Proof Induction in n, see PJE proposition (proof not given in class). Now we can prove that the product of primes guaranteed by Theorem 6.4 is unique (up to ordering).
3 Prime Numbers Theorem. Fundamental Theorem of Arithmetic (PJE theorem ) Every integer greater than 1 can be written as a product of primes unique up to ordering, i.e. for all n 2 there exist unique r and unique primes p 1 p 2 p r such that n = p 1 p 2... p r. Proof PJE p.283 Existence of prime factorisation is by Proposition 6.4. To prove uniqueness, assume for contradiction that two different products of primes give the same result. Cancel all primes which appear in both products, obtaining (*) p 1 p 2... p r = q 1 q 2... q s where p i q j for all i, j. Then p 1 q 1 q 2... q s so by Corollary 6.6 p 1 q j for some j. But since q j is prime, the only divisor of q j greater than one is q j. So p 1 = q j, contradicting (*). Primality testing: how to check if n is a prime number? From the definition of a prime number, it appears that we have to check each number from 2 to n 1 to see if it is a divisor of n. The process is made more efficient by the following result Proposition. (PJE proposition ) If m is composite then m has a prime divisor not exceeding m. That is, composite m, prime p: p m and p m. Proof PJE p.280 but I give an alternative proof here. Let m = p 1 p 2... p r be the prime factorisation of m with p 1 p 2 p r. Note that r 2 since m is not a prime. Then m p 1 p 1... p 1 = p r 1 p2 1. Hence m p Application: Sieve of Eratosthenes. The following algorithm will test whether N is prime. Write out the list of natural numbers from 2 up to N. Strike out all multiples of 2, except for 2. Strike out all multiples of the next remaining number except that number itself (this will be 3). Continue, at each step striking out all multiples of the next remaining number except that number itself. Stop when the next remaining number is > N.
4 Prime Numbers 54 Since we are striking out multiples we are only striking out composite numbers. Since every composite number m N has, by Proposition 6.8, a (prime) divisor m N, we will strike out every composite number N. Thus what will remain will be the noncomposite numbers, i.e. the primes, between 2 and N. So, for example, if we look for primes up to 100 we need only check for divisibility by primes up to 100 = 10, i.e. 2, 3, 5 and 7. See PJE example where this is done for N = 100. How many primes are there? Euclid was probably the first to prove, in his Elements, that there exist infinitely many prime numbers Theorem. (PJE theorem ) There are infinitely many primes. Proof. PJE p.285 seems to say that Euclid proved this by contradiction this is not the case. Euclid s proof is closer to the following. Let p 1,..., p r be any finite list of primes. We will show that there is a prime not in this list. Let N = p 1 p 2... p r + 1. By the Fundamental Theorem of Arithmetic, N has a prime divisor, say q. Note that N is not divisible by any of p 1,..., p r, since N leaves remainder 1 when divided by p i ; yet N is divisible by q. Hence q is a prime not equal to any of p 1,..., p r. Note It needs to be stressed that this N may well not be prime. The first few N are N = = 3, prime N = = 7, prime, N = = 31, prime N = = 211, prime N = = 2311, prime N = = = , composite. The point of the proof is that the prime divisors of N will be previously unseen primes. Fermat s Little Theorem and applications Lemma. If p is prime then p divides the binomial coefficient p 1. ( ) p for 1 r r
5 Proof. Recall that, whenever r is between 1 and p 1, Prime Numbers 55 ( ) ( ) p p 1 r = p r r 1 ( ) p (see Homework, Q10). Hence p r. But since 1 r < p, p does not divide r, so by r ( ) p Euclid s property of a prime (Theorem 6.5), p. r Remark. In your homework you were asked to show that n 5 n is divisibe by 5 for all n 1. The hint was to use induction and the Binomial Theorem on (k + 1) 5 which gives ( ) (k + 1) 5 = k 5 + 5k k k 2 + 5k + 1 k + 1 mod 5. The next Theorem generalises this Proposition. Freshman s Dream For all a, b Z and all primes p we have (a + b) p a p + b p mod p. p 1 ( ) p Proof By the Binomial Theorem, (a+b) p (a p +b p ) = a p r b r. By Lemma 6.11, r r=1 p divides every term on the right, hence p divides the whole sum, hence p (a + b) p (a p + b p ) as required. Now, just as ( ) can be used to prove n 5 n mod 5 for all n 1, Freshman s Dream can be used to prove the following general result Theorem. Fermat s Little Theorem (variant of PJE theorem ) For all n Z, and all primes p, n p n mod p. If gcd (p, n) = 1 then n p 1 1 mod p.
6 Prime Numbers 56 Proof. We fix a prime p. We first prove that n p n mod p for nonnegative n by induction in n. Base case: n = 0, 0 p 0 trivially. Inductive step: assume that k p k mod p. We need to prove (k + 1) p k + 1. Indeed, by Freshman s Dream, (k + 1) p k p + 1 mod p, which by inductive hypothesis is k + 1 mod p as required. So by induction n p n mod p for all n 0. Observe now that the result holds for all n Z: indeed, we proved that the congruence n p n mod p holds for n = 0, 1, 2,..., p 1 hence, by Modular Arithmetic, it holds for every n congruent, modulo p, to one of the numbers 0, 1, 2,..., p 1; but every integer is congruent, modulo p, to one of these residues. Now, to prove the second part of the theorem, write it as n n p 1 n 1 and assume that n is coprime to p. Then by Proposition 3.5 we can cancel n on both sides of the congruence to obtain n p 1 1 mod p Application: finding and using inverses. Fermat s Little Theorem has numerous applications, including simplifying the calculation of powers in modular arithmetic. From the theorem we see that if p is prime and p a then a p 1 1 mod p, or equivalently a p 2 a 1 mod p. This means that a p 2 is the inverse of a modulo p or, in the language of congruence classes, [a] 1 p = [ a p 2] p in Z p. Aside An interesting question is with which method is it quickest to calculate the modular inverse an integer modulo a prime. Is it by Euclid s Algorithm or by calculating a p 2 using, for example, the method of successive squaring? In fact, both methods have running time proportional to the number of decimal digits of a. For Euclid s algorithm this result on the running time is the content of Lame s Theorem, PJE p.226. End of aside We have seen before that a use of inverses is to solve linear congruences Example. Invert 5 mod 19. Hence solve the congruence 5x 6 mod 19. Solution 19 is prime so Fermat s Little theorem implies mod 19. So mod 19, i.e is the inverse of 5 mod 19. Thus multiplying both sides of the equation by 5 17 gives 5 18 x mod 19, i.e. x mod 19. Though this is an answer to the question we normally give x as the least positive residue. Successive squaring gives 5 2 6, , and mod 19.
7 Prime Numbers 57 Thus x mod 19, which should still be checked by substitution Application: Calculating powers. With p = 13 and a = 4 we see that mod 13. Thus = ( 4 12) 8 ( 4 2 ) ( 3) 2 9 mod 13, as we have seen before in Chapter 3, using the method of successive squaring. Aside Fermat s Little Theorem may appear wonderful in that it helps us solve congruences and simplifies substantially the calculation of large powers modulo p. But the result has one weakness, you need to know that the modulus is prime. It should be stressed that it is a difficult problem showing that a large number is prime (primality testing). End of aside Application: Fermat pseudoprimes. Testing whether n is a prime using Sieve of Eratosthenes may cost up to n operations. This is prohibitively expensive for large n. A approach is known as Fermat pseudoprime test, but unfortunately it is randomised and may give a false prime with nonzero probability: Pick a random integer a between 2 and n; Use Euclid s algorithm to find d = gcd(a, n). If d > 1, n is composite (divisible by d) and we are done; otherwise, calculate a n 1 mod n using successive squaring; if a n 1 1 mod n, then by Fermat s Little Theorem n is not prime. Unfortunately, if we discover that a n 1 1 mod n, we will not know if n is prime or composite. Indeed, a n 1 1 mod n holds for all prime n but holds also for some composite n and some a coprime to n. The number of steps in Euclid s Algorithm and successive squaring is only of order log n which is much smaller than n. But it can happen that, even after picking several a, no contradiction is found, but n is still composite. See your homework for an example.
8 Prime Numbers 58 Euler s phifunction and Euler s Theorem Recall from the previous chapter that Z n is the set of invertible classes in Z n and that it can be written as {[r] n : 0 r n 1, gcd (r, n) = 1} Definition. Euler s phifunction is, for all n 1, given by φ (n) = Z n = {0 r n 1 : gcd (r, n) = 1}. Examples Simply by checking we see that φ (5) = 4 and φ (7) = 6. In general φ (p) = p 1 for all primes p since all positive integers strictly less than p are coprime to p. Also by checking, we find that φ (8) = 4 and φ (16) = 8. In general φ (2 n ) = 2 n 1 for all n 1, since in these cases we are counting the integers coprime to 2 n, i.e. the odd integers. And half of the integers up to 2 n are odd. And you can easily check by hand that φ (6) = 2, φ (9) = 6 and φ (10) = Proposition. a) Multiplicativity of φ: If gcd(m, n) = 1 then φ(mn) = φ(m)φ(n). b) If p is prime then for all r 1 φ (p r ) = p r 1 (p 1). Proof Warning: although the result is designated here as a proposition, the proof of a) is one of the most difficult proofs in the course. Idea of proof of part a): the set of integers a such that 0 a m 1 and a is coprime to m has cardinality φ(m), see the definition of φ(m). Likewise, there are φ(n) integers b coprime to n such that 0 b n 1. Hence there are φ(m)φ(n) pairs (a, b) where a and b satisfy the given constraints. We show that such pairs (a, b) are in bijection with the set of integers x such that 0 x mn 1 and x is coprime to mn. Since the latter set has φ(mn) elements, the existence of a bijection implies that φ(m)φ(n) = φ(mn). Most of the proof below is a formal argument showing that there is indeed a bijection. Crucially, we use the Chinese Remainder Theorem which assumes that m and n are coprime. End of idea.
9 Prime Numbers 59 Proof of a). In this proof, we will use the following sets of nonnegative integers: M = {0, 1,..., m 1}, N = {0, 1,..., n 1}, MN = {0, 1,..., mn 1}. Assume that gcd(m, n) = 1. Then by Chinese Remainder Theorem 3.18, for each a M and b N there exists a unique integer in MN denote it x 0 (a, b) such that x 0 (a, b) a mod m and x 0 (a, b) b mod n. Hence we have a function x 0 : M N MN, (a, b) x 0 (a, b). The function x 0 is surjective: indeed, take x MN and let c the remainder of x modulo m (so that c M) and d be the remainder of x modulo n (so that d N). Then x c mod m and x d mod n, and x MN, which means that x = x 0 (c, d). Surjectivity of the function x 0 is proved. The set M N has cardinality m n, and the set MN has cardinality mn. A surjective function between two finite sets of the same cardinality is bijective. Hence the function x 0 is bijective. We will now restrict both the domain and the codomain of the function x 0 to obtain a bijective function between two nontrivial sets. The domain of the function x 0 is restricted to the set of pairs (a, b) M N such that a is coprime to m and b is coprime to n. The codomain of the function x 0 is restricted to the set of integers x MN such that x is coprime to mn. Of course we need to check that the function x 0 maps the new domain to the new codomain; that is, assuming that (a, b) M N is such that a is coprime to m and b is coprime to n, we need to show that x 0 (a, b) is coprime to mn. Indeed, since x 0 (a, b) a mod m and a is invertible mod m, we conclude that x 0 (a, b) is invertible mod m, i.e., is coprime to m. The same argument for b and n instead of a and m shows that x 0 (a, b) is coprime to n. Finally, if x 0 (a, b) is coprime to both m and n, then by homework question Q15, x 0 (a, b) is coprime to mn. Hence the resriction of x 0 is indeed a function (*) x 0 : {a : 0 a m 1, gcd(a, m) = 1} {b : 0 b n 1, gcd(b, n) = 1} {x : 0 x mn 1, gcd(x, mn) = 1}. Now, since the function x 0 is injective, its restriction (*) is injective. We claim that (*) is also surjective. To verify this claim, we assume that x = x 0 (a, b) is coprime to mn. We need to show that a is coprime to m and b is coprime to n. Indeed, if x is coprime to mn, then x is
10 Prime Numbers 60 coprime to m: any common divisor of x and m is also a common divisor of x and mn, the largest of which is 1. Hence x is invertible mod m, and since x = x 0 (a, b) a mod m, it follows that a is also invertible mod m, meanign that a is coprime to m. In the same way b is coprime to n. Thus, the function given by (*) is both injective and surjective, hence bijective. If a bijection exists between two sets, then the two sets have the same cardinality. It remains to observe that {a : 0 a m 1, gcd(a, m) = 1} {b : 0 b n 1, gcd(b, n) = 1} and = φ(m) φ(n) {x : 0 x mn 1, gcd(x, mn) = 1} = φ(mn). The two cardinalities are equal, so φ(m)φ(n) = φ(mn). b) See the homework questions for Week 12. Example. φ (100) = 40. Indeed, using Proposition 6.19, 100 = , 2 2 is coprime to 5 2 hence φ(100) = φ(2 2 )φ(5 2 ), φ(2 2 ) = = 2, φ(5 2 ) = = 20. The following is a generalization of Fermat s Little Theorem to composite moduli Theorem. Euler s Theorem If gcd (a, n) = 1 then a φ(n) 1 mod n. Proof is not in PJE s book, but the idea is to be found on p.290. Let us list the elements of Z n (i.e., invertible congruence classes mod n) as (List1) C 1, C 2,..., C φ(n) in some order. (Recall that Z n contains exactly φ(n) elements.) By Theorem 4.17, the product of elements of Z n is again an element of Z n, hence if X = C 1 C 2... C φ(n), then X Z n. By assumption, a is coprime to n, hence the residue class [a] n of a mod n belongs to the set Z n and is listed above as C i for some index i. Consider the following elements of Z n: (List2) C i C 1, C i C 2,..., C i C φ(n). This list is obtained by multiplying each entry of (List1) by C i.
11 Prime Numbers 61 We claim that all the entries in (List2) are distinct. Indeed, taking (List2) and multiplying all entries by Ci 1 gives (List1) which has φ(n) distinct entries. Hence (List2) also has φ(n) distinct entries, which means that the entries of (List2) are still all the elements of Z n, but possibly written in a different order. In particular, since multiplication on Z n is commutative (i.e., the product does not depend on the order in which the factors are written) the product of all the entries of (List2) is again X: X = (C i C 1 )(C i C 2 )... (C i C φ(n) ). Again using commutativity of multiplication, we rewrite this as X = C φ(n) i C 1 C 2... C φ(n). But this equation reads X = C φ(n) i X, which implies (multiplying through by X 1 ) that = [1] n. Recall that C i was the congruence class of a mod n, so that C φ(n) i [a φ(n) ] n = [1] n which merely means a φ(n) 1 mod n, as required Remark. Taking n = p, prime, in the Theorem we recover Fermat s Little Theorem. Indeed, φ(p) = p Example. Given φ (100) = 40 from above, find the last two digits in the decimal expansion of Solution We have to calculate mod 100 which is congruent to mod 100. Euler s Theorem tells us that 13 φ(100) = mod 100. Thus Now use successive squaring Hence = ( 13 40) mod , , , mod = mod 100. So the last two digits of are 77, as already found earlier.
12 Prime Numbers 62 Aside you may now think that Euler s Theorem has none of the problems of Fermat s Little Theorem as we do not need to know that the modulus m is prime; Euler s Theorem holds for all integers m. But unfortunately it has another problem, how to calculate φ (m)? In general this is very difficult for large m. Together, φ (st) = φ (s) φ (t) if gcd (s, t) = 1 and φ (p r ) = p r 1 (p 1) if p is prime, allow you to calculate φ (m) for any m provided you can factor m. Unfortunately, this is a very difficult problem for large m. Applications of Euler s and Fermat s Theorem. Not given in lectures Example. Find a solution to x 12 3 mod 11. Solution Any solution of this must satisfy gcd (x, 11) = 1 so Fermat s Little Theorem gives x 10 1 mod 11. Thus our equation becomes 3 x 12 x 2 x 10 x 2 mod 11. Now check. x x 2 mod Note that if x 6 then 11 x 5 and x 2 (11 x) 2 mod 11 so all possible values of x 2 mod 11 will be seen in the table. From the table we see an answer is x 5 mod Example. Show that x 5 3 mod 11 has no solutions. Solution by contradiction. Assume x 5 3 mod 11 has solutions. Any solution of this must satisfy gcd (x, 11) = 1 so Fermat s Little Theorem gives x 10 1 mod 11. Since 5 10 we square both sides of the original congruence to get 1 x 10 ( x 5) mod 11. This is false and so the assumption is false and thus the congruence has no solution.
13 Prime Numbers Example. Find a solution to x 7 3 mod 11. Solution Again x 10 1 mod 11 by Fermat s Little Theorem but this time 7 10, in fact gcd (7, 10) = 1. From Euclid s Algorithm we get (3) = 1. Raise both sides of the original congruence to the third power to get 3 3 ( x 7) 3 x 3 7 x by (3), x ( x 10) 2 x mod 11. Hence a solution is x mod 11. Don t forget to check your answer (by successive squaring of 5). iv) Is divisible by 11? Here we look at mod 11. Because 11 is prime we could use Fermat s Little Theorem to say mod 11. Thus = 33 0 mod 11, i.e is divisible by 11. Question for students. Show that is divisible by 65. Interesting problems concerning primes Not given in lectures. The following are conjectures and are all examples of problems that can be simply stated yet for which the answers are as yet unknown. 1) Goldbach s Conjecture, Is every even integer n 4 the sum of two primes? 4 = 2 + 2, 6 = 3 + 3, 8 = 3 + 5, 10 = 3 + 7, 12 = 5 + 7, 14 = , 16 = , 18 = , 20 = ,... Has been checked for all even numbers up to by Tomás Oliveira e Silva (as of 2017). Goldbach s Conjecture is a difficult problem because primes are multiplicative objects, defined in terms of divisibility, and yet this is an additive question. 2) Do there exist infinitely many Twin Primes, i.e. pairs of primes p and p such that p p = 2? The first few examples are (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73), (101, 103), (107, 109), (137, 139), (149, 151), (179, 181), (191, 193), (197, 199), (227, 229), (239, 241),....
14 Prime Numbers 64 A large prime pair is ± 1, with 388,342 decimal digits. It was discovered in September It has recently been proved (2014) that there are infinitely many pairs of primes p < p (not necessarily consecutive) with p p < ) Is n prime infinitely often? = 5, = 17, = 37, = 101, = 197,... It has been shown that n is either a prime or the product of two primes infinitely often. 4) For all m 1 does there exists a prime p : m 2 p (m + 1) 2? What is known is Theorem Bertrand s postulate: For all N 1 there exists a prime p : N p 2N. Proof. not given in course Definition. The Prime Counting Function If N Z, we define π(n) = {p Z : p N, p is prime}.
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