An integer p is prime if p > 1 and p has exactly two positive divisors, 1 and p.


 Rodney McCarthy
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1 Chapter 6 Prime Numbers Part VI of PJE. Definition and Fundamental Results Definition. (PJE definition ) An integer p is prime if p > 1 and p has exactly two positive divisors, 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. Lemma 6.1. An integer 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 is a positive divisor if n which is neither 1 nor n. Hence by definition, n is composite. Vice versa, if n is composite, then by definition 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 b = n/a. It remains to observe that 1 < b < n. Example. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19,... 56
2 Prime Numbers 57 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 77,232,917 1 with more than 23 million decimal digits, found in December 2017 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? Lemma 6.2. If p is prime, then for all integers a, either p a or p and a are coprime. Proof. PJE p.279. Let p be a prime and a be an integer. We assume that p a and prove that p is coprime to a. The only positive divisors of p are 1 and p. Out of these, p is not a divisor of a by assumption, so 1 is the only common positive divisor of p and a. Hence gcd(p, a) = 1, so p is coprime to a by definition of coprime. Proposition 6.3. (PJE proposition ) Every integer n 2 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. Inductive 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. Since k + 1 > 1, either k + 1 is a prime then it is a product of one prime, or k + 1 is composite then by Lemma 6.1, k + 1 = ab with 2 a, b k. By Inductive Hypothesis, a, b are products of primes, hence k + 1 = ab is also a product of primes. Conclusion: by strong induction, every integer n 2 is a product of primes. Theorem 6.4 (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. We assume that p ab and p a, and show that p b. By Lemma 6.2, p is coprime to a; but by assumption p ab, so by Corollary 2.5 p b, as required. Aside This result motivates the definition of a prime in more advanced work. End of aside
3 Prime Numbers 58 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 6.4, 6 is not prime. Corollary 6.5. 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 Proposition 6.3 is unique (up to ordering). Theorem 6.6 (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 6.3. 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. (*) Note that r, s > 0, i.e., there is at least one prime on each side of this equation, because a product of primes cannot be equal to 1. Then p 1 q 1 q 2... q s so by Corollary 6.5 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 may appear 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.
4 Prime Numbers 59 Proposition 6.7 (PJE proposition ). If m is composite then m has a prime divisor not greater than 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 1. Application: Sieve of Eratosthenes The following method will test whether N is prime, and will generate all the prime numbers between 1 and N. Write out the list of natural numbers from 2 up to N. Strike out all multiples of 2, except for 2 itself. 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 greater than N. Since we are striking out multiples we are only striking out composite numbers. Since every composite number m N has, by Proposition 6.7, 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, the following statement. Theorem 6.8 (PJE theorem ). There are infinitely many primes.
5 Prime Numbers 60 Proof. PJE p.285. Assume for contradiction that there are finitely many primes, and list all primes as p 1,..., p r. Let N = p 1 p 2... p r + 1. Observe that N is not divisible by any prime, since N leaves remainder 1 when divided by p i for all i. Yet by the Fundamental Theorem of Arithmetic, N must be divisible by a prime, a contradiction. 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 6.9. If p is prime then p divides the binomial coefficient Proof. Observe that, whenever r is between 1 and p 1, ( ) ( ) p p 1 r = p. r r 1 ( ) p for 1 r p 1. r r p! (p r)! r! = Indeed, by the Factorial Formula for the binomial coefficient, this translates as ( ) p (p 1)! p (p r)! (r 1)! which is true. Hence p r. But since 1 r < p, p does not divide r, so by r ( ) p Euclid s property of a prime (Theorem 6.4), p. r 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 6.9, p r r=1 divides every term on the right, hence p divides the whole sum, hence p (a + b) p (a p + b p ) as required.
6 Prime Numbers 61 Theorem Fermat s Little Theorem (variant of PJE theorem ) For all n Z, and all primes p, If gcd (p, n) = 1 then n p n mod p. n p 1 1 mod p. 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 3.3 we can divide both sides of the congruence by n 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?
7 Prime Numbers 62 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. 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.
8 Prime Numbers 63 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. 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 φ: N N given, for all n 1, by φ(n) = Z n = {r : 0 r n 1, gcd(r, n) = 1}. Example. Simply by checking we see that φ (5) = 4 and φ (7) = 6.
9 Prime Numbers 64 In general φ (p) = p 1 for all primes p since all positive integers strictly less than p are coprime to p by Lemma 6.2. 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) = 4. 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. a) See the homework questions for Week 12. Not examinable. b) We claim that the numbers not coprime to p r are multiples of p. Indeed, if x is not coprime to p r, that is, gcd(x, p r ) > 1, then gcd(x, p r ) is divisible by a prime, but the only prime which divides p r is p, hence p x. Vice versa, if p x then x is not coprime to p r. Among the first p r nonnegative integers there are pr p integers divisible by p, therefore φ(p r ) = {0, 1,..., p r 1} \ {x : gcd(x, p r ) > 1} = p r {0, p, 2p,... } = p r p r 1. Corollary. If n = p k 1 1 pk pks s = s distinct primes, then i=1 p k i i φ(n) = is the prime factorisation of n, where p 1,..., p s are s i=1 p k i 1 i (p i 1). Proof. If s = 1, this is just part b) of the Proposition. Otherwise, p k 1 1 is coprime to s ( i=2 s ) φ(n) = φ(p k 1 1 )φ = p k (p 1 1) s i=2 p k i i i=2 use induction) to obtain the required formula. p k i i p k i i. Continue in the same way (more formally: Example. φ (100) = 40. Indeed, using Proposition 6.12, 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. so
10 Prime Numbers 65 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. Claim. If gcd(a, n) = 1, then the function f : Z n Z n defined by the rule f ([x] n ) = [a] n [x] n is bijective. Proof of the claim: gcd(a, n) = 1 implies that the class [a] n Z n is invertible. Consider the function g : Z n Z n defined by g([x] n ) = [a] 1 n [x] n. Then f(g([x] n )) = g(f([x] n )) = [x] n for all [x] n Z n, hence g is the inverse function to f. A function which has an inverse is a bijection. The claim is proved. To prove Euler s Theorem, denote by X the product of all elements of Z n. X = [x] n Z n If we apply f to each factor in this product, we will get the same factors but perhaps in a [x] n. different order, because f is a bijection. Hence the product will not change: X = Therefore, by definition of the function f, X = [x] n Z n [x] n Z n f ([x] n ) ([a] n [x] n ) = [a] Z n n [x] n Z n Since X Z n, X has an inverse in Z n. Multiply through by X 1 : [x] n = [a] Z n n X. [a] Z n n = [1] n which is equivalent to a Z n 1 mod n. Theorem is proved. Since Z n = φ(n) by definition of φ(n), Euler s Remark. Taking n = p, prime, in the Theorem we recover Fermat s Little Theorem. Indeed, φ(p) = p 1.
11 Prime Numbers 66 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. 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.
12 Prime Numbers 67 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 11. 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. 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 = 1. (6.1) Raise both sides of the original congruence to the third power to get 3 3 ( x 7) 3 x 3 7 x by (6.1), x ( x 10) 2 x mod 11. Hence a solution is x mod 11.
13 Prime Numbers 68 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),.... 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 < 246.
14 Prime Numbers 69 3) 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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