Wilson s Theorem and Fermat s Theorem
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1 Wilson s Theorem and Fermat s Theorem Wilson s theorem says that p is prime if and only if (p 1)! = 1 (mod p). Fermat s theorem says that if p is prime and p a, then a p 1 = 1 (mod p). Wilson s theorem and Fermat s theorem can be used to reduce large numbers with respect to a give modulus and to solve congruences. They are also used to prove other results in number theory for example, those used in cryptographic applications. Lemma. Let p be a prime and let 0 < k < p. k 2 = 1 (mod p) if and only if k = 1 or k = p 1. Proof. If k = 1, then k 2 = 1 (mod p). If k = p 1, then Conversely, suppose k 2 = 1 (mod p). Then k 2 = p 2 2p + 1 = 1 (mod p). p k 2 1 = (k 1)(k + 1), and since p is prime, p k 1 or p k + 1. The only number in {1,..., p 1} which satisfies p k 1 is 1, and the only number in {1,..., p 1} which satisfies p k + 1 is p 1. Theorem. (Wilson s theorem) Let p > 1. p is prime if and only if (p 1)! = 1 (mod p). Proof. Suppose p is prime. If k {1,..., p 1}, then k is relatively prime to p. So there are integers a and b such that ak + bp = 1, or ak = 1 (mod p). Reducing a mod p, I may assume a {1,..., p 1}. Thus, every element of {1,..., p 1} has a reciprocal mod p in this set. The preceding lemma shows that only 1 and p 1 are their own reciprocals. Thus, the elements 2,..., p 2 must pair up into pairs {x, x 1 }. It follows that their product is 1. Hence, (p 1)! = 1 2 (p 2) (p 1) = 1 1 (p 1) = p 1 = 1 (mod p). Now suppose (p 1)! = 1 (mod p). I want to show p is prime. Begin by rewriting the equation as (p 1)! + 1 = kp. Suppose p = ab. I may take 1 a, b p. If a = p, the factorization is trivial, so suppose a < p. Then a (p 1)! (since it s one of {1,..., p 1}) and a p, so (p 1)! + 1 = kp shows a 1. Therefore, a = 1. This proves that the only factorization of p is the trivial one, so p is prime. Example. Wilson s theorem implies that the product of any ten consecutive numbers, none divisible by 11, equals 1 mod 11 (since any ten consecutive numbers reduce mod 11 to {1, 2,..., 10}. For example, = 1 (mod 11). 1
2 Example. Find the least nonnegative residue of 70! (mod 5183). Note that 5183 = I ll start by finding the residues of 70! mod 71 and 73. By Wilson s theorem, 70! = 1 (mod 71). Next, let k = 70! (mod 73). Then Note that 2 37 = 74 = 1 (mod 73). So Thus, k = 70! (mod 73), ( 2)( 1)k = 72! (mod 73), 2k = 1 (mod 73). 37 2k = 37 ( 1) (mod 73), k = 37 = 36 (mod 73). 70! = 1 (mod 71) and 70! = 36 (mod 73). I ll the the iterative method of the Chinese Remainder Theorem to get a congruence mod First, 70! = 1 (mod 71) means 70! = a for some a Z. Plugging this into the second congruence yields a = 36 (mod 73), 71a = 37 (mod 73), 2a = 37 (mod 73), ( 37)( 2a) = ( 37)(37) (mod 73), a = 1369 = 18 (mod 73). The last congruence means that a = b for some b Z. Plugging this into 70! = a gives 70! = ( b) = b, or 70! = 1277 (mod 5183). Theorem. (Fermat s Theorem) Let p be prime, and suppose p a. Then a p 1 = 1 (mod p). Proof. The idea is to show that the integers a, 2a,..., (p 1)a reduce mod p to the standard system of residues {1,..., p 1}, then apply Wilson s theorem. There are p 1 numbers in the set {a, 2a,..., (p 1)a}. So all I need to do is show that they re distinct mod p. Suppose that 1 j, k p 1, and aj = ak (mod p). This means p aj ak = a(j k), so p a or p j k. Since the first case is ruled out by assumption, p j k. But since 1 j, k p 1, this is only possible if j = k. Thus, {a, 2a,..., (p 1)a} are p 1 distinct numbers mod p. So if I reduce mod p, I must get the numbers in {1,..., p 1}. Hence, a 2a (p 1)a = 1 2 (p 1) = (p 1)! = 1 (mod p). 2
3 On the other hand, another application of Wilson s theorem shows that a 2a (p 1)a = a p 1 (p 1)! = a p 1 (mod p). So a p 1 = 1 (mod p), or a p 1 = 1 (mod p). Corollary. If p is prime, then a p = a (mod p) for all a. Proof. If p a, then a p = 0 (mod p) and a = 0 (mod p), so a p = a (mod p). If p a, then a p 1 = 1 (mod p). Multiplying by a, I get a p = a (mod p) again. Example. Compute (mod 83). One way is to multiply out ; Mathematica tells me it is Now just reduce mod 83. Heh. If you don t have Mathematica, maybe you should use Fermat s theorem , so Fermat says = 1 (mod 83). Now 3 82 = 246, so = = (50 82 ) = = 100 = 17 (mod 83). In other words, if you re trying to reduce a k mod p, where p a, factor out as many a p 1 s as possible, then reduce the rest by hand. Example. Solve 16x = 25 (mod 41). I d like to multiply both sides by the reciprocal of 16 mod 41. What is it? Well, I could use the Euclidean algorithm on (16, 41), or I could do a multiplication table mod 41. A simpler approach is to note that by Fermat, = 1 (mod 41). Hence, x = (mod 41) gives x = (mod 41). Now this is an answer, but a rather cheesy one. I ought to reduce the right side mod 41 to something a little smaller! I can t use Fermat any more, so I just divide and conquer = 256 = 10 (mod 41), so Now 10 2 = 100 = 18 (mod 41), so 18 2 = 324 = 37 (mod 41), so = (16 2 ) 19 (16 25) = = (mod 41) = (10 2 ) 9 (10 31) = = (mod 41) = (18 2 ) 4 (18 23) = = = 10 4 = 40 (mod 41). (I reduce down to the point where the arithmetic can be handled by whatever computational tools I have available.) c 2006 by Bruce Ikenaga 3
4 Pseudoprimes and Mersenne primes If p is prime, the little Fermat theorem implies that 2 p =2 (modp) : The ancient Chinese thought for empirical reasons that the converse was true: p j 2 p ; 2 implies p is prime. For instance, 2 13 ; 2 = 8190 = and 13 is indeed prime. Unfortunately, the result is not true in general. Example. Since 11 6 j 2, little Fermat implies 2 10 = 1 (mod 11). Therefore, Next, 2 31 =2 (mod31),so = 1 (mod 11) = 2 (mod 11) : (2 31 ) 11 =2 11 =2 (2 5 ) 2 = =2 1 2 = 2 (mod 31) : That is, = 2 (mod 31). Now 11 and 31 are prime, so = 2 (mod 341). However, 341 = isn't prime is Imagine trying to do this without congruences! Denition. Let n be a positive composite number, b>0. n is a pseudoprime to the base b if b n = b (mod n) : In other words, n j b n ; b. The Chinese case above was b =2. Example. 341 is a pseduprime to the base 2. Example. (Mersenne primes) The k-th Mersenne number is M k =2 k ; 1: If M k is prime, it is a Mersenne prime. Claim: If M k is prime, then k is prime. Suppose M k is prime and k = ab, wherea b > 1. Now 2 k ; 1=2 ab ; 1=(2 a ) b ; 1=(2 a ; 1) ; (2 a ) b;1 +(2 a ) b; a +1 : This is a proper factorization of M k. This contradiction establishes the claim. 1
5 Mersenne thought the converse was true: If k is prime, then M k is prime. However, 2 11 ; 1=2047=23 89: Hence, Mersenne's conjecture is false. However, ; 1 is prime, so you can sometimes come up with large primes this way. Proposition. If k is prime, then M k is a pseudoprime to the base 2. Proof. Since M k =2 k ; 1, 2 k =1 (modm k ). Suppose k is prime. Then 2 k = 2 (mod k) k j 2 k ; 2 2 k ; 2=kj for some j: Then (2 k ) j =1 (modm k ) 2 kj =1 (modm k ) 2 2k ;2 =1 (modm k ) 2 2k ;1 =2 (modm k ) 2 Mk =2 (modm k ) : This proves that M k is a pseudoprime to the base 2. Question. Are there innitely many Mersenne primes? Note that the even perfect numbers are exactly the numbers of the form 2 p;1 (2 p ; 1), where 2 p ; 1 is a Mersenne prime. So the existence of innitely many Mersenne primes would imply the existence of innitely many even perfect numbers. c1996 by Bruce Ikenaga 2
6 Euler s Theorem If n is a positive integer, φ(n) is the number of integers in the range {1,..., n} which are relatively prime to n. φ is called the Euler phi-function. Euler s theorem generalizes Fermat s theorem to the case where the modulus is not prime. It says that if n is a positive integer and (a, n) = 1, then a φ(n) = 1 (mod n). Question: How can you generalize the little Fermat theorem to the case where the modulus is composite? Idea: The key point of the proof of Fermat s theorem was that if p is prime, {1, 2,..., p 1} are relatively prime to p. This suggests that in the general case, it might be useful to look at the numbers less than the modulus n which are relatively prime to n. This motivates the following definition. Definition. The Euler φ-function is the function on positive integers defined by φ(n) = (the number of integers in {1, 2,..., n 1} which are relatively prime to n). Example. φ(24) = 8, because there are eight positive integers less than 24 which are relatively prime to 24: 1, 5, 7, 11,13,17,19,23 On the other hand, φ(11) = 10, because all of the numbers in {1,..., 10} are relatively prime to 11. Here s a graph of φ(n): You can see that the function jumps around a little, but the data points are bounded above by the line y = x. A point will be nearly on this line whenever n is prime, and since there are infinitely many primes, there will always be points near it. Later, I ll derive a formula for computing φ(n) in terms of the prime factorization of n. Remarks. 1
7 1. If p is prime, φ(p) = p 1. This is clear, because all of the numbers {1,..., p 1} are relatively prime to p. 2. φ(n) counts the elements in {1, 2,..., n 1} which are invertible mod n. For (a, n) = 1 if and only if ax = 1 (mod n) for some x. (For people who know some abstract algebra, φ(n) is the order of the group of units Z n.) Definition. A reduced residue system mod n is a set of numbers such that: a 1, a 2,..., a φ(n) (a) If i j, then a i a j (mod n). That is, the a s are distinct mod n. (b) For each i, (a i, n) = 1. That is, all the a s are relatively prime to n. Thus, a reduced residue system contains exactly one representative for each number relatively prime to n. Compare this to a complete residue system mod n, which contains exactly one representative to every number mod n. Example. {1, 5, 7, 11} is a reduced residue system mod 12. So if { 11, 17, 31, 1}. On the other hand, {0, 1, 2, 3,4,5,6, 7, 8,9,10,11} is a complete residue system mod 12. Lemma. Let φ(n) = k, and let {a 1,..., a k } be a reduced residue system mod n. (a) For all m, {a 1 + mn,..., a k + mn} is a reduced residue system mod n. (b) If (m, n) = 1, {ma 1,..., ma k } is a reduced residue system mod n. Proof. (a) This is clear, since a i = a i + mn (mod n) for all i. (b) Since (m, n) = 1, I may find x such that mx = 1 (mod n). Since (a i, n) = 1, so I may find b i such that a i b i = 1 (mod n). Then (xb i )(am i ) = (mx)(a i b i ) = 1 (mod n), which proves that am i is invertible mod n. Hence, (am i, n) = 1 the ma s are relatively prime to n. Now if ma i = ma j (mod n), then xma i = xma j (mod n), or a i = a j (mod n). Since the a s were distinct mod n, this is only possible of i = j. Hence, the ma s are also distinct mod n. Therefore, {ma 1,..., ma k } is a reduced residue system mod n. Corollary. Let φ(n) = k, and let {a 1,..., a k } be a reduced residue system mod n. Suppose (s, n) = 1, and let t be any integer. Then {sa 1 + tn, sa 2 + tn,..., sa k + tn} is a reduced residue system mod n. Example. {1, 5} is a reduced residue system mod 6. Adding 12 = 2 6 to each number, I get {13, 17}, another reduced residue system mod 6. Since (6, 25) = 1, I may multiply the original system by 25 to obtain {25, 125}, another reduced residue system. Finally, { , } = {37, 137} is yet another reduced residue system mod 12. 2
8 Theorem. (Euler s Theorem) Let n > 0, (a, n) = 1. Then a φ(n) = 1 (mod n). Remark. If n is prime, then φ(n) = n 1, and Euler s theorem says a n 1 = 1 (mod n): the little Fermat theorem. Proof. Let φ(n) = k, and let {a 1,..., a k } be a reduced residue system mod n. I may assume that the a i s lie in the range {1,..., n 1}. Since (a, n) = 1, {aa 1,..., aa k } is another reduced residue system mod n. Since this is the same set of numbers mod n as the original system, the two systems must have the same product mod n: (aa 1 ) (aa k ) = a 1 a k (mod n), a k (a 1 a k ) = a 1 a k (mod n). Now each a i is invertible mod n, so multiplying both sides by a 1 1 a 1 k, I get a k = 1 (mod n), or a φ(n) = 1 (mod n). Example. φ(40) = 16, and (9, 40) = 1. Hence, 9 16 = 1 (mod 40) surely not an obvious fact! Likewise, = 1 (mod 40). You can also use Euler s theorem to compute modular powers. Suppose I want to find (mod 40). Mathematica tells me that is I probably don t want to do this by hand! Euler s theorem says that = 1 (mod 40). So = = (33 16 ) = 9 2 = 81 = 1 (mod 40). Example. Solve 15x = 7 (mod 32). Note that (15, 32) = 1 and φ(32) = 16. Therefore, = 1 (mod 32). Multiply the equation by : x = (mod 32). Now = = 105 (15 2 ) 7 = = = 9 (mod 32). So the solution is x = 9 (mod 32). c 2006 by Bruce Ikenaga 3
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