12. Let Rm = {0,1,2,..., m 1} be a complete residue system modulo ra. Let a be an integer. When is a Rm = {0,1 a, 2 a,...

Transcription

1 12. Let Rm = {0,1,2,..., m 1} be a complete residue system modulo ra. Let a be an integer. When is a Rm = {0,1 a, 2 a,..., a (ra - 1)} a complete residue system modulo m? Prove your conjecture. (Try m = 15 and various values of a.) 13. Show that if p is prime, then the simultaneous linear congruence ax + by = u (mod p) ex + dy = v (mod p) has a unique solution a:, y modulo p when ad bc^ 0 (mod p). 14. Let a, 6, and ra be integers such that (a, m) = (6, m) = 1. Determine the number of pairs (x,?/) with x and?/ in a complete residue system, satisfying the congruence ax+by = 0 (mod m). What can you say about the number of solutions if the conditions (a, ra) = 1 or (6, ra) = 1 are dropped? 15. (Clement, 1949) Let n> 2. (a) Show that 4[(n - 1)! + 1] + n = 0 (mod n) if and only if n is prime. (b) Show that 2(n - 1)! = -1 (mod n + 2) if and only if n + 2 is prime. Hence show that 4[(n - 1)! + 1] + n = 0 (mod n + 2) if and only if (c) n + 2 is prime. Show that the integers n and n + 2 form a pair of twin primes if and only if 4[(n-l)! + l]+n = 0 (mod [n(n + 2)]). 3.3 Chinese Remainder Theorem An important problem is to find integers satisfying many different divisibil ity conditions. Also, in many applications, we reduce a computation modulo composite numbers to a computation over its prime factors. The Chinese Re mainder Theorem is the fundamental tool that allows us to combine congru ences to reach conclusions about the original problem. The idea first appears in the writings of the Chinese mathematician Sun-Tzu, who lived in the third century. The method was further developed by Chin Chiu-Shao in the thir teenth century. In the West, Euler seems to have been the first to study it extensively. The method is fundamental and has numerous applications. We begin with a motivating example. Example Determine the smallest positive integer that gives a remain der of 2 upon division by 3, a remainder of 1 upon division by 5, and a remainder of 6 upon division by 7. Let a: be a solution; then the conditions require that x = 2 (mod 3) (1) x = 1 (mod 5) (2) x = 6 (mod 7). (3)

2 This is a system of three congruences on x. We can solve it as follows. The first congruence implies that x = 3k + 2 for some A:. Using this in equation (2) gives a condition on k. 3k+ 2= 1 (mod 5), 3k = 1-2 = 4 (mod 5). (4) We can solve for k by multiplying by 3"1 mod 5, which is 2. Hence k = 2-4 = 8 = 3 (mod 5). Hence there exists an integer r such that k = 5r + 3, so x = 15r Use this value of x in (3) to obtain 15r + ll = 6 (mod 7) 15r = 6-11 = 2 (mod 7). Since 15 = 1 (mod 7), we have r = 2 (mod 7). Therefore, there exists an integer 5 so that r = 7s + 2; hence x = 15(75 + 2) + 11 = It can be verified that every x of this form is a solution. The smallest positive solution is x = 41, and it is the only solution in the interval 0 < x < 105. Observe that 105 is the product of 3, 5, and 7, the three moduli. Exercise. Find the smallest positive solution to the following simultaneous congruence. x = 3 (mod 7) x = 8 (mod 11). Let us apply the technique of the above example to find the general solu tion in the case of two simultaneous congruences. Suppose mi and rri2 are two relatively prime integers. Given a\ and a^, we want to find x so that x = ai (mod rrti), i = 1,2. The first congruence implies x = m\v + a\ for some r. Using this in the second congruence gives m\r + a\ = (22 (mod or m\t = d2 ai (mod 7712) Let u be an inverse of m\ modulo Then r = u(a2 a\) (mod 7712); that is, there exists an integer s so that r = ra25 + u(a2 «i); hence x = mirri2s + m\ua2 m\uai + a\. Instead of starting with the first equation (with respect to mi), we could have started with the second and obtained a different form for the solution. Since the solution is the same, we can write it in a more symmetric form. Because, m\u = 1 (mod 7712), we can write 1 m\u = vknri2 for some v. Therefore x = mim2s + m\ua2 + rri2va\. Now, x = a\ (mod mi) implies that 77^ = 1 (mod mi); that is, v is the inverse of m2 (mod mi).

3 The numbers u and v can be found by the extended Euclidean Algorithm. If miu + miv = 1, then we have shown that every solution to the two congru ences satisfies x = um\a2 + vrri2ai (mod raira2). Conversely, it is easy to show that any such x is a solution to the congruence. The following exercise extends this technique to three equations. The reader is urged to try the fol lowing exercise before reading the proof of the Chinese Remainder Theorem. Exercise. Let mi, 777,2, and m% be integers that are relatively prime in pairs. Suppose u and v are integers such that urrti + vm,2 = 1. (a) Show that three congruences x = a^ (mod m^), i = 1,2,3 can be reduced to the two congruences x = um\a2 + vm20i\ x = as (mod 7713). (mod Now, apply the technique for the solution of two congruences to this pair. Let y and z be integers such that ym\m,2 + zms 1. Show that every solution to the simultaneous congruence is of the form x = ymirri2as + uzm\mza2 + vzrri2m^ai (mod ). (b) Show that y is the inverse of mi777,2 (mod 777,3), uz is the inverse of 777,17713 (mod 7712), and vz is the inverse of ,3 (mod mi). (c) Use this to find the general solution in Example (d) Do you see a pattern in the general solution to two and three congru ences? Generalize to an arbitrary number of congruences. Chinese Remainder Theorem. Let mi, 777,2,., wv be pairwise relatively prime integers. Then the simultaneous congruence x = a\ (mod 777,1) x = CL2 (mod 7712) x = ar (mod mr) has a unique solution modulo the product mi777,2 * * * ^r- PROOF. The computations performed above for two and three simultaneous congruences show the general pattern. We give an explicit formula for the solution and then show its uniqueness.

4 Let M = mirri2 mr, and M{ = M/mi. Since the rrti s are pairwise relatively prime, M* is relatively prime to m^, so it has an inverse X{ modulo rrii\ that is, MiXi = 1 (mod mi). If i ^ j, then mi \ Mf, that is, Mj = 0 (mod mi). Then x = a\m\x\ + (12M2X2 H h armrxr (mod M) is a solution to the system because a\m\x\ + (I2M2X2 H h armrxr = aimixi (mod m*) = a^ 1 (mod m^) = a^ (mod m^). Suppose a: and y are two distinct solutions. Now x = y (mod mi) for each i; that is, rrii \ (x y). The m^ are pairwise relatively prime, hence, by unique factorization, the product m\m2 -mr \ (x y)\ that is, x = y (mod mi mr). This proves the uniqueness. The virtue of the proof is that it gives an explicit formula for the solution. Example Let's solve the congruence in Example using the for mula given in the proof of the theorem. First, we have to compute the quantities M». From their definition, M = = 105, Mi = 35, M2 = 21, and M3 = 15. We can now determine the inverses, either by the Euclidean Algorithm or by inspection, as the numbers involved are small. The inverses X{ of M{ are: a;i = 2 (mod 3), x2 = 1 (mod 5), x3 = 1 (mod 7). Therefore, x = (mod 105) = (mod 105) = 251 (mod 105) = 41. (mod 105). We verify: 41 = 2 (mod 3), 41 = 1 (mod 5) and 41 = 6 (mod 7). Since the solution is unique modulo the product, all other solutions are of the form x = &, where k is an integer. Exercise. Find the smallest positive number satisfying the following congru ences. x = 3 (mod 8) x = 2 (mod 5) x = 6 (mod 9).

5 Suppose m = p^... pekk is the prime factorization of m. The Chinese Remainder Theorem states that any a, 1 < a < ra, is completely determined by knowing the remainders n upon division by p\{; that is, a is the unique solution, satisfying 1 < a < ra, to the set of congruences x = r* (mod p^) for i = 1,..., k. This is the most useful feature of the Chinese Remainder Theorem. A natural question is to try to extend the Chinese Remainder Theorem to moduli that are not necessarily relatively prime. Example Consider the congruences x = 3 (mod 8) x = 7 (mod 12). As 12 is not a prime power, we can break the second congruence into x = 7 (mod 3) and x = 7 (mod 4); that is, x = 1 (mod 3) and x = 3 (mod 4). We now have three congruences x = 3 (mod 8) x = 7 (mod 4) # = 7 (mod 3). Now, x = 3 (mod 8) implies x = 3 (mod 4); hence x = 3 (mod 4) is not necessary. We have two remaining congruences x = 3 (mod 8) x = 1 (mod 3). Now the moduli are relatively prime and we can solve this pair of congru ences using the previous technique to obtain x = 19 solution. (mod 24) as the unique We can also solve the congruences by successive substitution; the con gruence x = 3 (mod 8) means that x = 3 + 8k for some integer k. Sub stitute this into the second equation to obtain 3 + 8k = 7 (mod 12), or 8/c = 4 (mod 12). As (8,12) = 4 and 4 4, we obtain 2k = 1 (mod 3). As 2 is invertible modulo 3, k = 2 (mod 3); that is, fc = 2 + 3r and x = 3 + 8(2 + 3r) = r for some integer r. Exercise. Does the congruence x = 5 (mod 8) z = 7 (mod 12) have a solution?

6 Ch'in Chiu-Shao (c ) N Ch'in Chiu-Shao was born in 1202 in the province of Szechwan. He seems to have been extremely bright and excelled in many fields. He served in the military and later as a Governor, a position in which he is supposed to have accumulated immense wealth. In 1247, he published the Shu-Shu Chiu-Chang ("nine sections of math ematics"). The work includes his solution to simultaneous congruences, a method essentially the same as what we use today. In addition, Ch'in Chiu-Shao was the first to find a method to approximate roots of polyno mial equations. This is the same method published by Horner in 1819 and vis now known as Horner's method. > The following theorem generalizes the Chinese Remainder Theorem to arbitrary moduli. Theorem Let m\,..., mr be integers; then the system of congruences x = di (mod mi), i = 1,..., r has a solution if and only if for all i ^ j, (mj, rrij) di dj. The solution is unique modulo [mi,..., mr]. PROOF. It suffices to prove the theorem for two integers mi and First, consider the case where mi = pei for some prime p. For i 7^ j, if ej > eu then x = dj (mod pei) implies that x = dj (mod pei)\ on the other hand, x = di (mod pei)\ therefore, ai = dj (modpe*), or pei \ di dj. [Here e; = min(ei, ej), so {pe\pej) = pei.] In general, let mi = p\l -pekk and rri2 = pfx -pdkk. Then x = a\ (mod mi) is equivalent to x = a\ (mod p^) for i = 1,..., k, and x = d2 (mod 7722) is equivalent to x = d<i (mod pf) for i = 1,..., fe. Looking at the congruence modulo powers of p$, we see that ai = a2 (mod p m^clj ^), or equivalently, ai = d2 (mod (mi,m2)). This is the necessary condition for the set of simultaneous congruences to have a solution. Conversely, if (mi, raj) di dj for all i ^ j9 then we can solve the congruences by reducing to the case of relatively prime moduli. If ei > di, then d\ = d<i (mod (mi, 7712)) and x = a\ (modp?*) imply that x = a2 (mod pf), and we can drop the second congruence. (Similarly, if e* < di.) Hence we have a set of congruences x = di where a» is ai or d<i depending on where the maximum occurs. Now we can apply the Chinese Remainder Theorem to obtain a unique solution modulo the LCM.

7 Example Consider the system x = 5 (mod 8) x = 7 (mod 14) x = 21 (mod 35). Now, (8,14) (7-5) and (35,14) (21-7), so the system has a unique solution modulo [8,14,35] = 280. To solve the system, we reduce to looking at prime powers: (1) (2) (3) (4) (5) Now, (1) implies (2), so we can drop (2). Equations (3) and (5) are the same; therefore, we have a system of three congruences, x = 5 (mod 8), x = 0 (mod 7), x = 1 (mod 5), with relatively prime moduli. Remark. The Chinese Remainder Theorem can also be proved by the fol lowing algebraic interpretation. This proof shows the existence of a solution without giving any method for finding one. Recall that Z/raZ is the set of equivalence classes of integers that are congruent modulo ra. Let (ra, n) = 1, consider the product Z/raZ x Z/nZ consisting of pairs (x,y) with x G Z/raZ and y G Z/nZ. We can define a function from Z/ranZ to Z/raZ x Z/nZ as follows: if a G Z/ranZ: define f(a) = (a mod ra, a mod n). It can be easily checked that / preserves addition and multiplication; that is, f(ab) = f(a)f(b). If f(a) = 0, that is, a = 0 (mod m) and a = 0 (mod n), then a is divisible by ran, that is, a = 0 (mod ran). This is equivalent to the fact that / is one-to-one. Now, Z/ranZ has ran elements, and Z/raZ x Z/nZ also has

8 ran elements. Since the number of elements is the same, the function / must also be onto. Equivalently, / is an isomorphism (of rings) Z/ranZ «Z/mZ x Z/nZ. The fact that / is onto is equivalent to the Chinese Remainder Theorem be cause if we are given a G Z/raZ and b G Z/nZ, then there is a unique x G Z/mnZ such that a = x (mod m) and b = x (mod n). Exercises for Section Solve the following simultaneous congruences. (a) x = l (mod 2) (b) x = 7 (mod 9) x = 2 (mod 3) x = 0 (mod 10) # = 4 (mod 5) x = 3 (mod 7). a? = 2 (mod 7). 2. Determine if the following simultaneous congruences have a solution, and find the smallest positive solution if it exists. (a) x = 3 (mod 8) (b) z = 4 (mod 6) x = 7 (mod 12) x = 8 (mod 12) ar = 4 (mod 15). x = 12 (mod 18). 3. Write a computer program to solve a set of simultaneous congruences when the moduli are relatively prime. You can use the Extended Euclidean Algo rithm to find the solution for two moduli and then use this inductively for an arbitrary set of moduli. This program will be used frequently in later sections, so test it well. 4. [Bhaskara] There are n eggs in a basket. If eggs are removed from the basket 2, 3, 4, 5, and 6 at a time, there remain 1, 2, 3, 4, and 5 eggs in the basket respectively. If eggs are removed from the basket 7 at a time, no eggs remain in the basket. What is the smallest possible number of eggs the basket could have contained? 5. [Ch'in Chiu-Shao] Three farmers divide equally the rice that they have grown. One goes to a market where an 83-pound weight is used, another to a market that uses a 110-pound weight, and the third to a market using a 135-pound weight. Each farmer sells as many full measures as possible, and when the three return home, the first has 32 pounds of rice left, the sec ond 70 pounds, and the third 30 pounds. Find the total amount of rice they took to the market. 6. You are asked to design a system for numbering TV programs to facilitate the programming of VCRs. Each program should be assigned a number so