7.2 Applications of Euler s and Fermat s Theorem.

Transcription

1 7.2 Applications of Euler s and Fermat s Theorem. i) Finding and using inverses. From Fermat s Little 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. An interesting question is with which method is it quickest to calculate the modular inverse. 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. Aside III And in general, for (a, m) = 1 the inverse of a modulo m is a φ(m) 1. But now it isn t reasonable to calculate a φ(m) 1 since it is difficult to evaluate φ (m) for composite m. In fact we need to know that m is prime to use φ (m) = m 1, and it is another difficult task to verify whether a given integer is prime or not. The largest known prime (as of September 2008) is 2 43,112, (See known prime for further details). A use of inverses is to solve linear congruences. Example Find the integer solutions of 5x 6 mod 19. Solution From Euler s theorem mod 19 (because φ (19) = 18 since 19 is prime). 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 agrees with the answer found in Chapter 3, but which should still be checked by substitution. ii) 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. 1

2 iii) Solving non-linear congruences. Example Solve x 12 3 mod 11. Solution Any solution of this must satisfy gcd (x, 11) = 1 so Fermat s Theorem gives x 10 1 mod 11. Thus our equation becomes x 2 3 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 the 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 Theorem gives x 10 1 mod 11. Since 5 10 we square both sides of the 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 Solve x 7 3 mod 11. Solution Again x 10 1 mod 11 but this time 7 10, in fact gcd (7, 10) = 1. From Euclid s Algorithm we get = 1. Raise both sides of the congruence to the third power to get 3 3 ( x 7) 3 x 3 7 x x ( x 10) 2 x mod 11. Hence the solution is x mod 11. Don t forget to check your answer (by successive squaring of 5). 2

3 iv) Last decimal digits Example Find the last two digits in the decimal expansion of Solution We have to calculate mod 100. Euler s Theorem tells us that 13 φ(100) = mod 10. Thus Now use successive squaring = ( 13 40) mod , , , mod 100. (1) Hence mod 100. So the last two digits of are 77, as already found in Chapter 3. This method makes it easier to find the last two digits of , also asked for in Chapter = ( 13 40) mod 100. Hence, using (1), mod 100. Therefore, the last two digits of are 49. What are the last three digits of ? See ap- Question for students. pendix. 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. 3

4 Appendix Contents 1. Sieve of Eratosthenes 2. Prime Number Theorem 3. Lowest Common Multiples 4. Use of unique factorization 5. What are φ (100) and φ (1000) and their use 6. Further examples of Fermat s Theorem 7. Wilson s Theorem 8. Permutations on Z 8 1) Sieve of Eratosthenes. All the numbers from 2 up to Delete multiples of 2 apart from 2 itself:

5 Delete multiples of 3 apart from 3 itself: Delete multiples of 5 apart from 5 itself: Delete multiples of 7 apart from 7 itself: The next number is 11 which is greater than 100 and so there are no multiples of it less than 100 that haven t already been deleted in earlier stages. Thus we are left with the 25 primes <100. 5

6 2) Prime Numbers. For x > 0 let π (x) be the number of primes not exceeding x. So π (10) = 4, π (100) = 25, (as seen from the application of the Sieve of Eratosthenes in the appendix), π (1000) = 168 and π (5000) = 669. Also π ( 10 23) = 1, 925, 320, 391, 606, 803, 968, 923, due to Tomás Oliveira e Silva, Is there a simple formula for π (x)? Theorem Prime Number Theorem (1896) π (x) x/ ln x 1 as x. This means that we can make π (x) / (x/ ln x) 1 as small as we like by taking x sufficiently large. This difference is < if x 10 9, is < if x and is < if x So, for very large x the graph for π (x) lies close to that of x/ ln x. Proof not given. You might in fact think that x/ ln x is not a good approximation to π (x). If f (x) = x/ ln x then f ( 10 23) = 1, 888, 236, 877, 840, 225, 337, which seems quite a long way short of the true value of π (10 23 ) above. (See function for further details including a description of a better approximation to π (x).) 3) Lowest Common Multiples. Let a, b be integers > 1, and let p 1, p 2,..., p n be all the primes that divide ab. Then a = n for some a i, b i 0 for 1 i n. Theorem and gcd (a, b) = n lcm (a, b) = n p a i i, b = n p b i i, p d i i with d i = min (a i, b i ) for all 1 i n, p f i i with f i = max (a i, b i ) for all 1 i n. 6

7 And gcd (a, b) lcm (a, b) = ab. Proof The result for the gcd (a, b) was given in lectures. For the lowest common multiple, recall the Definition The lowest common multiple of integers a, b is the positive integer f that satisfies (1) a f, b f, (2) if a k, b k then f k. Note that ab is a multiple of both a and b and thus a common multiple. By part (2) of the definition lcm (a, b) ab. In particular the primes dividing lcm (a, b) come from the list p 1, p 2,..., p n. Therefore for some f i 0 for 1 i n. lcm (a, b) = n The condition that a lcm (a, b) implies a i f i while b lcm (a, b) implies b i f i for 1 i n. These combine to give max (a i, b i ) f i for 1 i n. But lcm (a, b) is the least common multiple so we take f i = max (a i, b i ) for 1 i n. Finally, since for all x, y we have (student to check this), then p f i i min (x, y) + max (x, y) = x + y, gcd (a, b) lcm (a, b) = n = n p min(a i,b i ) i n p max(a i,b i ) i p min(a i,b i )+max(a i,b i ) i = n p a i+b i = n = ab. n p a i p b i 7

8 4) Example of use of unique factorization. Example For any m 2 there is no rational solution to q m = 2. Solution by contradiction. Assume that for some m 2 there exists a rational q : q m = 2. Write q = a/b with a, b Z, so we get a m = 2b m. First this means a m > 1 and so a > 1 in which case it has a decomposition into primes. Thus ( n n ) m p ma i i = = 2b m. By unique factorization, one of the primes in the product has to be 2, assume p 1 = 2 so a 1 1. Now either b = 1 or it is a product of primes, one of which may be 2. So 2 ma 1 n i=2 p a i i p ma i i = 2 1+mb 1 where b 1, and all other b i 0. By unique factorization the number of 2 s on both sides are identical so ma 1 = 1 + mb 1, i.e. m (a 1 b 1 ) = 1 in which case m divides 1. This contradicts m 2 and so the assumption is false and thus for no m 2 can we find a rational solution of q m = 2. n i=2 p mb i i One of the first proofs you examine at University is to prove that 2 is irrational, i.e. no rational solutions of q 2 = 2. So here we have extended this result. 5) What is φ (100)? Lemma For all s Z, gcd (r + s10, 100) = 1 gcd (r, 10) = 1. Proof ( ) Assume gcd (r, 10) = 1. Assume for contradiction that l ±1 such that l 100 and l (r + s10). Then since 100 is not prime, l 100 implies there exists d ±1 such that d l and d 10. But l (r + s10) then implies that d (r + s10). Combining d 10 and d (r + s10) gives d r. We thus have d 10 and d r which with gcd (r, 10) = 1 means that d is either d = 1 or d = 1. This contradicts d ±1, so the assumption is false, thus the only common divisors of r + s10 and 100 are ±1. ( ) The contrapositive of what we want to show, namely gcd (r, 10) > 1 gcd (r + s10, 100) > 1 8

9 is obvious. Corollary φ (100) = 40. Proof Start from φ (100) = {1 n 100 : gcd (n, 100) = 1} = {1 r 10, 0 s 9 : gcd (r + s10, 100) = 1} = {1 r 10, 0 s 9 : gcd (r, 10) = 1} by the lemma = 10 {1 r 10 : gcd (r, 10) = 1} = 10φ (10) = 40. Example For m 1 φ (10 m ) = 10φ ( 10 m 1). Solution Simply write every 1 n 10 m as r + s10 n 1 with 1 r 10 n 1 and 0 s 9. Then the same reasoning as before gives gcd ( r + s10 m 1, 10 m) = 1 gcd ( r, 10 m 1) = 1, for any s Z. And again, as before, this implies φ (10 m ) = 10φ (10 m 1 ). Example Find the last three digits of Solution We need calculate mod From above φ (1000) = 10φ (100) = 400, and so mod Thus Repeated squaring gives ( ) mod = 169, = mod 1000, = mod 1000, = mod 1000, = mod 1000, = mod 1000, = mod

10 Combine = = = mod Hence the last 3 digits of are ) Further examples of the use of Euler s and Fermat s Theorems. Example Show that is divisible by 65. Solution We need show that 65 ( ). Since 65 = 5 13 we need show that 5 ( ) and 13 ( ). First, since φ (5) = 4 we have mod 5. Hence = ( 2 4) = 5 0 mod 5. Secondly, since φ (13) = 12 we have mod 5. Hence = ( 2 12) = 65 0 mod 13. Combining these we get the required result. Example Is 221 prime? Solution Fermat s Little Theorem tells us that If 221 is prime then mod 221. Note that 220 = =

11 Look at powers of 2 modulo 221. So n 2 2n = ( 2 2n 1 ) 2 mod = = = = ( 101) 2 = ( 101) = ( 101) ( 101) mod 221. Since mod 221 we deduce that 221 is not prime. Example You now notice that 221 is composite and in fact 221 = Use Fermat s Little Theorem, and not the method of successive squaring modulo 221, to check that mod 221. Solution. If x mod (17 13) then x mod 17 and x mod 13. By Fermat s theorem we have mod 17 so = ( 2 4) 3 ( 1) mod 17. Similarly mod 13 so = = 16 3 mod 13. Thus our two equations become x 16 mod 17 and x 3 mod 13 11

12 Such a system was solved in the Appendix to Chapter 3, using the Chinese Remainder Theorem, where we found x 16 mod 221. Example Solve x 22 + x 11 2 mod 11. Solution Any solution must have gcd (x, 11) = 1 and so, by By Fermat s Theorem, x 22 + x 11 x 2 + x x x (x + 6) 2 36 mod 11, by completing the square. Thus we need only solve (x + 6) mod 11. From the table x x 2 mod we see that x mod 11 and x mod 11 are solutions, i.e. x 1 or 9 mod 11. Example Show that there are no integer solutions (x, y) to x 12 11x 6 y 5 + y Solution We assume there are integer solutions. modulo 11 they remain solutions. There are two cases. When we look at this First, it maybe that 11 y in which case the equation becomes x 12 8 mod 11. For any solution of this we must have gcd (x, 11) = 1 so, again by Fermat s Theorem, x 10 1 mod 11 and so we get x 2 8 mod 11. From the table above we see this has no solutions. Secondly, perhaps 11 y, i.e. gcd (y, 11) = 1. So Fermat s Theorem again gives both x 10, y 10 1 mod 11. Thus x 12 11x 6 y 5 + y 10 x x mod 11, and so we are looking for solutions to x 2 7 mod 11. Again from the table we see this has no solution. In both cases our equation has no solutions modulo 11. This contradiction means our original equation has no integer solutions. 12

13 7) Wilson s Theorem. Recall that Z m = {[r] m : 1 r m, (r, m) = 1} = {[r] m : 1 r m, [x] m Z m : [r] m [x] m = [1] m }. Question What 1 r m are self-inverse modulo m, i.e. for which we can we take [x] m = [r] m in [r] m [x] m = [1] m? In other words, for which 1 r m do we have r 2 1 mod m? Answer given here only for m = p, prime. Theorem x 2 1 mod p if, and only if, x 1 or 1 mod p. Proof x 2 1 mod p p ( x 2 1 ) p (x 1) (x + 1) p (x 1) or p (x + 1) since p prime x 1 mod p or x 1 mod p. Thus the only self-inverses in Z p are [1] p and [p 1] p. As a corollary of this we have Theorem Wilson s Theorem. If p is prime then (p 1)! 1 mod p. Proof p.291. Take the product of all the classes in Z p: 1 r p 1 (r,p)=1 [r] p. Rearrange, pairing up a class with its inverse, leaving [1] p and [p 1] p unpaired. So the product becomes ) [1] p ( pairs [r] p [r] 1 p [p 1] p = [p 1] p. Thus 1 r p 1 (r,p)=1 [r] p = [p 1] p, 13

14 which is equivalent to the stated result. Example Calculate 20! mod 23. Solution 23 is a prime so Wilson s Theorem gives 22! 1 mod 23. But 22! = ! ( 1) ( 2) 20! 2 20! mod 23. By observation 12 is the inverse of 2 modulo 23 so 20! ! 12 22! mod 12. 8) Permutations on Z 8 Example On Z 8 = {1, 3, 5, 7}, (dropping the [...] 8 ), we have four such permutations. Note that ρ 1 = ρ 3 = ρ 5 = ρ 7 = ( ) = Z 8, ( ) = (1, 3) (5, 7), ( ) = (1, 5) (3, 7), ( ) = (1, 7) (3, 5) ρ b ρ a ([r] 8 ) = ρ b (ρ a ([r] 8 )) = ρ b ([ar] 8 ) = [b (ar)] 8 = [(ba) r] 8 = ρ ba ([r] 8 ). Hence ρ b ρ a = ρ ba. Thus we can draw up the multiplication table for these permutations which can be compared with that for (Z 8, ) : ρ 1 ρ 3 ρ 5 ρ 7 ρ 1 ρ 1 ρ 3 ρ 5 ρ 7 ρ 3 ρ 3 ρ 1 ρ 7 ρ 5 ρ 5 ρ 5 ρ 7 ρ 1 ρ 3 ρ 7 ρ 7 ρ 5 ρ 3 ρ The mapping Z 8 {ρ 1, ρ 3, ρ 5, ρ 7 }, r ρ r preserves the structure of the table. Perhaps you can generalize this to a result for (Z m, )? 14