MATH10040 Chapter 3: Congruences and the Chinese Remainder Theorem

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1 MATH10040 Chapter 3: Congruences and the Chinese Remainder Theorem 1. Congruence modulo m Recall that R m (a) denotes the remainder of a on division by m. Thus, by the division algorithm, 0 R m (a) < m and a = mt+r m (a) for some t Z. Lemma 1.1. Fix an integer m 1. Let a, b Z. The following are equivalent: (1) m a b (2) a = b + ms for some s Z (3) R m (a) = R m (b) Proof. (1) (2): m a b if and only if there exists s Z with a b = ms if and only if a = b + ms for some s Z. (2)= (3): Suppose that a = b + ms for some s Z. Now, b = mt + R m (b) for some t Z. Thus a = m(t + s) + R m (b). By the uniqueness part of the division algorithm, R m (b) = R m (a). (3)= (2): Suppose that R m (a) = R m (b). Call this number r. Then a = mt+r and b = ms+r for some s, t Z. It follows that a b = m(t s) is a multiple of m. Definition 1.2. We say that a is congruent to b modulo m and write a b (mod m) if m a b (i.e. if any of the equivalent conditions of the Lemma hold). Here the number m is called the modulus of the congruence. Example (mod 7) (mod 7) (mod 7) (mod 7). Example 1.4. Which numbers are congruent to 13 modulo 6? Answer: a 13 (mod 6) if and only if a = t for some t Z; i.e. if and only if a is one of the numbers..., 13 12, 13 6, 13, , , ,... So a 13 (mod 6) if and only if a is one of the numbers..., 1, 7, 13, 19, 25, 31,... (Note: Since 13 1 (mod 6), this is also the list of numbers which are congruent to 1 modulo 6; i.e. this is the list of numbers which leave remainder 1 on division by 6.) 1

2 2 Remark 1.5. Observe that if m 1 and a Z, then the set of numbers which are congruent to a modulo m are the numbers in the list..., a 2m, a m, a, a + m, a + 2m, a + 3m,... This is a (doubly) infinite sequence of numbers. The difference between successive terms is always m. (Thus this is an arithmetic progression with common difference m.) Remark 1.6. Of course, we always have that a R m (a) (mod m) since a = R m (a) + mt for some t Z. Thus R m (a) is the unique number satisfying 0 R m (a) < m and a R m (a) (mod m). Remark 1.7. Observe that, from the definition, m a if and only if a 0 (mod m). We will use this repeatedly below; problems about divisibility can be rephrased as problems about congruences and vice versa. What turns the notion of congruence into a very powerful mathematical tool is the following crucial Theorem: Theorem 1.8 (Algebra of Congruences). Let m 1. (1) If a b (mod m) and b c (mod m) then a c (mod m). (2) If a a (mod m) and b b (mod m) then a + b a + b (mod m). (3) If a a (mod m) and b b (mod m) then ab a b (mod m). Proof. (4) If a a (mod m) then a n (a ) n (mod m) for any n 1. (1) R m (a) = R m (b) and R m (b) = R m (c) = R m (a) = R m (c). (2) m a a and m b b = m (a a )+(b b ) = (a+b) (a +b ) = a + b a + b (mod m). (3) We have a = a + mt, b = b + ms for some s, t Z. But then ab = (a + mt)(b + ms) = a b + m(a s + b t + mst) so that ab a b (mod m).

3 3 (4) We ll use induction on n 1. The case n = 1 is clear. Suppose the result is known for n. Then a n (a ) n (mod m) and a a (mod m) = a n a (a ) n a (mod m) by (3) = a n+1 (a ) n+1 (mod m). This theorem says that when doing algebra modulo m, at any stage we can replace a number a with another (usually smaller) number a whenever a a (mod m). 2. Algebra of congruences: Examples Example 2.1. Calculate the remainder of on division by 11. Solution: 14 3 (mod 11) and 37 4 (mod 11). Therefore, using properties (3) and (1) of the algebra of congruences, (mod 11). Therefore the remainder is 1. Example 2.2. Calculate the remainder of on division by 17. Solution: 35 1 (mod 17) = (mod 17) (using properties (1) and (4) of the algebra of congruences). So the answer is 1. Example 2.3. Calculate the remainder of on division by 13. Solution: (mod 13). Thus ( 1) 49 1 (mod 13). Since 1 12 (mod 13), (mod 13), so the remainder is 12. Example 2.4. Calculate the remainder of on division by 7. Solution: First, 16 2 (mod 7) = (mod 7). The strategy now is to find a small power of 2 which is congruent to ±1 modulo 7: In this case, we have (mod 7). It follows that for any integer m, 2 3m = (2 3 ) m 1 m 1 (mod 7); i.e. if n is a multiple of 3 then 2 n 1 (mod 7). Now we wish to calculate 2 55 (mod 7). 55 is not a multiple of 3. But 55 = and 54 is a multiple of 3. Thus 2 55 = = (mod 7). So the answer is 2.

4 4 Example 2.5. Find the remainder of on division by 11. Solution: We first calculate (mod 11) and (mod 11) using the strategy from the last example: : First 14 3 (mod 11). Now look at small powers of 3 to find one congruent to ±1 modulo = 9 2 (mod 11). Thus (3 2 ) 5 ( 2) (mod 11); i.e (mod 11). It follows (as in the last example) that 3 n 1 (mod 11) whenever n is a multiple of 10. So (mod 11) : (mod 11). Now (mod 11) = 4 5 = (2 5 ) 2 ( 1) 2 1 (mod 11). So 4 n 1 (mod 11) whenever n is divisible by 5. So (mod 11). Finally, using property (2) of the algebra of congruences, (mod 11). So the answer is 8. Example 2.6. Show that for every n 1 the number 2 2n + 5 is composite. Solution: We look at a few examples for small n to see if we can make any useful observations or conjectures: n = = 9 = 3 3 n = = 21 = 3 7 n = = 261 = After this the numbers grow too fast. However, based on our evidence so far we might guess that these numbers are always divisible by 3 (and hence are not prime). Thus, our guess is 3 2 2n + 5 for all n 1. This can be rewritten 2 2n (mod 3) or 2 2n 5 1 (mod 3). So we must prove that 2 2n 1 (mod 3) for all n: But 2 1 (mod 3) = 2 2n ( 1) 2n 1 (mod 3) for all n (since 2 n is even). Congruences often give us a means to prove that an equation has no integer solutions: Example 2.7. Show that the equation x 2 = 3y + 2 has no integer solutions x and y.

5 Solution: Note that numbers of the form 3y + 2 are precisely the numbers which are congruent to 2 modulo 3. Thus we are required to show that the congruence x 2 2 (mod 3) has no solutions. Now x leaves either remainder 0, 1 or 2 on division by 3. So x 0, 1 or 2 (mod 3). We consider each of these possibilities in turn: If x 0 (mod 3) then x (mod 3), and so x 2 2 (mod 3). If x 1 (mod 3) then x 2 1 (mod 3), and so x 2 2 (mod 3). If x 2 (mod 3) then x (mod 3), and so x 2 2 (mod 3). Example 2.8. Show that no number of the form 4t + 3 (t Z) can be written as a sum of two squares. Solution: The numbers of the form 4t + 3 are precisely the numbers congruent to 3 modulo 4. So we are required to prove that the congruence x 2 + y 2 3 (mod 4) has no solutions. Now, for n Z, n 0, 1, 2 or 3 (mod 4) and so n 2 0, 1, 0, 1 (mod 4). So any square is congruent either to 0 or 1 modulo 4. Thus x 2, y 2 0 or 1 (mod 4) and hence x 2 + y , 0 + 1, 1 + 0, (mod 4). Thus x 2 + y 2 0, 1 or 2 (mod 4), and hence x 2 + y 2 3 (mod 4). Example 2.9. Show that the equation x 2 2y 2 = 10 has no integer solutions. Solution: The strategy here is the following. Note that if a, b Z then a = b = a b (mod m) for any modulus m: To prove that a b it is enough to find a modulus m for which a b (mod m). Thus we attempt to find a modulus m for which x 2 2y 2 10 (mod m) has no solutions. There is no golden rule for choosing such a modulus; in general it may require some experimentation. However, one possibility worth exploring is the divisors of the numbers of coefficients occurring in the equation. Since 10 = 2 5 it is worth trying m = 2 or m = 5. If we try m = 2, we obtain the congruence x 2 0 (mod 2) which has solutions (take x to be any even number), so this is inconclusive. If we try m = 5 we get the congruence x 2 2y 2 0 (mod 5). Now if n is any integer then n 0, 1, 2, 3 or 4 (mod 5) and hence n 2 0, 1, 4 0, 1, 1 (mod 5). If 5 x, y then x 2, y 2 ±1 (mod 5) and x 2 2y 2 (±1) 2(±1) 0 (mod 5). This shows there are no solutions to the congruence (and hence to the original equation) satisfying 5 x, y. 5

6 6 So if there is a solution, we must have 5 x or 5 y, and hence 5 x and 5 y (since 5 10). Modulo 5 we then get 0 0 (mod 5), which does not tell us anything about the solvability of the original equation. However, observe that if 5 x, y then 5 2 x 2, y 2 and hence 25 = 5 2 x 2 2y 2. Since the equation x 2 2y 2 = 10 has no solutions in this case either Divisibility by 9. Let N be a natural number. Suppose that the (standard) decimal i.e. base 10 representation of N has the form a k a k 1 a 0 (where a 0,..., a k {0, 1,..., 9} are decimal digits). Then, by definition, N = a a a k a k. (i.e. this is meaning of the decimal representation). (For example, if the decimal representation of N is , then N = ) Since 10 1 (mod 9), 10 n 1 (mod 9) for any n. Thus N = a a a k a k a 0 + a a k (mod 9). Example What is the remainder of on division by 9? Solution: (mod 9). So the answer is 3. Exercise If N has decinal representation a k a 0 show that N a 0 a ( 1) k a k (mod 11). 3. Decomposing moduli Lemma 3.1. Suppose that m, n are nonzero integers satisfying (m, n) = 1. Suppose a is any integer. Then mn a m a and n a. Proof. = : If mn a then a = mnt = m(nt) = n(mt) for some integer t, and thus a is a multiple of m and of n. =: Since m a we have a = mb for some integer b. Since n mb and (n, m) = 1 we have n b. Thus b = nc for some integer c and hence a = mb = mnc. Remark 3.2. Of course, the hypothesis (m, n) = 1 is crucial here. For example 6 60 and 4 60 but The best that can be said in general is that if m a and n a then the least common multiple of m and n must divide a (see last problem on Homework 2.)

7 Corollary 3.3. Suppose that m, n are positive integers satisying (m, n) = 1 and that a, b Z. Then a b (mod mn) a b (mod m) and a b (mod n). Proof. a b (mod mn) mn (a b) m (a b) and n (a b) (by Lemma 3.1) a b (mod m) and a b (mod n). Example 3.4. Show that (mod 20). Since 20 = 4 5 and (4, 5) = 1, by Corollary 3.3, it is enough to show that (mod 4) and (mod 5). Now 87 1 (mod 4) = ( 1) (mod 4) (mod 5) = (mod 5). Thus (mod 5). Example 3.5. Find the last two digits of We wish to calculate the remainder of on division by 100; i.e. what is (mod 100)? Since 100 = 4 25 (and (4, 25) = 1) we can look at (mod 4) and (mod 25) separately (mod 4): 57 1 (mod 4) = (mod 4) (mod 25): 57 7 (mod 25). Now (mod 25) and hence (mod 25). Thus (mod 25). Let r be the remainder of on division by 100. We ve shown r 1 (mod 4) and r 18 (mod 25). So we just have to find the number smaller than 100 which satisfies both of these conditions. From the second condition r {18, 43, 68, 93}. Of these only 93 is congruent to 1 modulo 4. So r = 93. We can generalize this useful result to the situation where the modulus is a product of three or more (relatively prime) factors: First we need the following observation. Lemma 3.6. Suppose that m 1,..., m t, m are nonzero integers with (m, m i ) = 1 for i = 1,..., t. Then (m, m 1 m t ) = 1. Proof. Suppose FTSOC that (m, m 1 m t ) > 1. Then there is a prime number p satisfying p m and p m 1 m t. Since p is prime, p m i for some i and hence (m, m i ) 1, a contradiction. Lemma 3.7. Suppose that m 1,..., m t are positive integers satisfying (m i, m j ) = 1 whenever i j (we say that the list of numbers is pairwise relatively prime). Let m = m 1 m t. 7

8 8 If a, b Z, then a b (mod m) a b (mod m i ) for all i = 1,..., t. Proof. = is clear. =: We ll proceed by induction on t 2. When t = 2 this is just Corollary 3.3. Suppose the result is known for t, and that we are given m 1,..., m t, m t+1 pairwise relatively prime. Let m = m 1 m t m t+1. Since (m t+1, m i ) = 1 for i = 1,..., t it follows that (m t+1, m 1 m t ) = 1 by Lemma 3.6. Now m = (m 1 m t ) m t+1 and therefore a b (mod m) a b (mod m 1 m t ) and a b (mod m t+1 )(by the case t = 2) a b (mod m i ) for all i t and a b (mod m t+1 ) (by ind. hyp.) a b (mod m i ) for all i t + 1. Corollary 3.8. Let m > 1 and suppose that m = p a 1 1 p at t factorization of m. If a, b Z we have is the prime a b (mod m) a b (mod p a i i ) for all i = 1,..., t. Example 3.9. Let n be any integer not divisible by 2, 3 or 5. Prove that n 4 1 (mod 30) (i.e. 30 n 4 1). Solution: Since 30 = we just have to prove that n 4 1 (mod m) for m = 2, 3 and 5. Now n 1 (mod 2) = n 4 1 (mod 2). Also n ±1 (mod 3) = n 4 (±1) 4 1 (mod 3). Finally n ±1, ±2 (mod 5). So n 4 (±1) 4 or (±2) 4 (mod 5) = n 4 1 (mod 5). Remark Lemma 3.7 requires that the moduli are pairwise relatively prime. For example, the three numbers 6, 10, 15 satisfy (6, 10, 15) = 1 but no pair of them is relatively prime. Not that 31 1 (mod 6), 31 1 (mod 10) and 31 1 (mod 15) but 31 1 (mod ). So the conclusion of Lemma 3.7 does not hold in this case.

9 9 4. Solving ax b (mod m) Fix a modulus m and let a, b be any integers. We say that s Z is a solution of the congruence ax b (mod m) if as b (mod m). We want to determine when such a congruence has a solution, and, if so, to find all solutions. Note that as b (mod m) as = b mt for some t Z as + mt = b for some t Z. Thus the congruence ax b (mod m) has a solution s if and only if the equation ax + my = b has solutions s and t. But we have already seen (in Chapter 2, section 4) how to find all integer solutions of ax + my = b: (1) The equation has integer solutions if and only if (a, m) b. (2) Let g = (a, m) and suppose that g b. Then the equation ax+my = b has the same solutions as the equation a g x + m g y = b g. Therefore, ax b (mod m) has the same solutions as a g x b g (mod m g ). (3) Dividing across by g if necessary (as in (2)) we can reduce to the case (a, m) = 1: Using Euclid s algorithm or otherwise we find s, t Z with as+mt = b (and thus as b (mod m)). The general solution of the equation ax + my = b is x = s + mk, y = t ak, k Z. Therefore the general solution the congruence ax b (mod m) is x = s + mk, k Z; i.e. x s (mod m). (4) To conclude: When (a, m) = 1, first find a solution s of ax b (mod m) (using Euclid s algorithm or otherwise). The general solution is then the set of all integers congruent to s modulo m (we call this the congruence class of s modulo m).

10 10 Example 4.1. Find all integer solutions of the congruence 111x 9 (mod 69). Solution: First (111, 69) = 3 and 3 9, so the congruence has solutions. We divide across by 3 to get the simpler congruence 37x 3 (mod 23) (which has exactly the same solutions by our arguments above). By Euclid s algorithm we find 1 = ( 8) and so 3 = ( 24). Thus s = 15 is a solution of the congruence. The general solution is x = k, k Z (i.e. the congruence class of 15mod23). 5. The Chinese Remainder Theorem Example 5.1 (From a thirteenth century Chinese manuscript). Three farmers divide their rice equally. One goes to the market where an 83kg weight is used, another to a market where a 110kg weight is used and the third to a market where a 135kg weight is used. Each farmer sells as many full measures as they can. On returning, there remains 32, 70 and 30 kg respectively. Q: What is the total amount of rice? Let x denote the amount of rice each farmer has. So the total amount is 3x. So we know x 32 (mod 83) x 70 (mod 110) x 30 (mod 135) (Caution: The moduli are not pairwise relatively prime here: (110, 135) 1.) The Chinese Remainder Theorem tells us how to solve a system of congruences like the one in this example. Theorem 5.2 (Chinese Remainder Theorem). Suppose that m 1,..., m t are positive and pairwise relatively prime. Let m = m 1 m t. Let a 1,..., a t Z.

11 11 (1) There exists c Z satisfying c a 1 (mod m 1 ), c a 2 (mod m 2 ),..., c a t (mod m t ) (2) If c is one solution then the general solution is x = c + ms, s Z. Proof. (1) For i = 1, 2,..., t let n i = m/m i. So m = m i n i. (1) Note that (m i, n i ) = 1 for all i by Lemma 3.6, since the numbers m 1,..., m t are pairwise relatively prime. Therefore, for each i the congruence n i x 1 (mod m i ) is solvable; i.e. for each i there exists an integer b i satisfying n i b i 1 (mod m i ). (2) On the other hand if j i then n j b j 0 (mod m i ) since m i n j. Now let c := a 1 n 1 b a t n t b t. Fix some i. By congruence (2) a j n j b j 0 (mod m i ) whenever j i and hence c a i n i b i (mod m i ). However, by congruence (1) n i b i 1 (mod m i ) and hence c a i (mod m i ) as required. (2) Suppose that d is any other solution of the system of congruences. Then c d (mod m i ) i = c d (mod m)(by Lemma 3.7) so that d = c + ms for some s. Conversely, if d = c+ms for some s, then d c (mod m) and hence, for each i, d c a i (mod m i ) so that d is a solution. Remark 5.3. Note that in the proof of the first part of the theorem we provide a recipe for calculating c. Example 5.4. Find all solutions of the system of simultaneous congruences x 3 (mod 5), x 5 (mod 7), x 7 (mod 11). Solution: We use the recipe described in the proof of the CRT. Here a 1 = 3, m 1 = 5, a 2 = 5, m 2 = 7, a 3 = 7 and m 3 = 11. So n 1 = 7 11 = 77, n 2 = 5 11 = 55, n 3 = 5 7 = 35.

12 12 To find b 1, solve 77x 1 (mod 5); i.e. 2x 1 (mod 5). So we can take b 1 = 3. To find b 2, solve 55x 1 (mod 7); i.e. x 1 (mod 7) (since 55 1 (mod 7)). So we can take b 2 = 1. To find b 3, solve 35 1 (mod 11); i.e. 2x 1 (mod 11). So we can take b 3 = 6. Then a solution of our simultaneous congruences is c = a 1 n 1 b 1 + a 2 n 2 b 2 + a 3 n 3 b 3 = ( 1) = = Now m = m 1 m 2 m 3 = = 385. So the general solution x = t, t Z. This is the congruence class of 1888 modulo 385. Since (mod 385), this is the same as x = t, t Z Note that it follows 348 is the smallest positive solution of the simultaneous congruences; i.e. 348 is the smallest positive number which leaves remainder 3 on division by 5, 5 on division by 7 and 7 on division by 11. Example 5.5. We ll solve the thirteenth century Chinese problem above. The problem is to find the smallest positive integer solution of the simultaneous congruences x 32 (mod 83), x 70 (mod 110), x 30 (mod 135). We begin by recalling that the moduli in this problem are not pairwise relatively prime since (110, 135) = 5 so we cannot yet apply the procedure of the last example. First, we have to decompose the moduli. congruence is equivalent to x 0 (mod 5) and x 4 (mod 22) and the third congruence is equivalent to x 0 (mod 5) and x 3 (mod 27). So the original system is equivalent to the system Since 110 = 5 22 the second x 32 (mod 83), x 4 (mod 22), x 0 (mod 5) and x 3 (mod 27) and hence (recombining the last two) to the system x 32 (mod 83), x 4 (mod 22), and x 30 (mod 135)

13 13 So we now apply the method of the proof: To find b 1 : Solve x 1 (mod 83). Now modulo 83 we have (mod 83). So we must solve 65x 1 (mod 83). We use Euclid s algorithm: which gives Thus we can take b 1 = = = = = = = = ( 18). To find b 2 : We must solve x 1 (mod 22). Now (mod 22). So the congruence is 7x 1 (mod 22). We can take b 2 = 3. To find b 3, we must solve the congruence 83 22x 1 (mod 135). Now (mod 135). So we must solve 71x 1 (mod 135). Applying Euclid s algorithm to the pair 71, 315 we find the solution b 3 = 19. Thus a solution of the system of congruences is c = ( 3) ( 19) = Now = , so the general solution is x = t, t Z; i.e. the set of all numbers congruent to modulo Now the remainder of on division by is So is the smallest positive solution of the system. Therefore each farmer had kg of rice and the total amount of rice was = kg. Example 5.6. Find all positive integers less than 5000 which leave remainders 2, 4, 8 when divided by 9, 10, 11 respectively. Solution: First we find the general solution of the system x 2 (mod 9), x 4 (mod 10), x 8 (mod 11). To find b 1, solve 110x 1 (mod 9); i.e. 2x 1 (mod 9). We can take b 1 = 5.

14 14 To find b 2, we solve 99x 1 (mod 10); i.e. x 1 (mod 10). So we let b 2 = 1. To find b 3, we solve 90x 1 (mod 11); i.e. 2x 1 (mod 11). So let b 3 = 6. This gives the solution c = ( 1) = Since m = = 990, and since (mod 990) the general solution is the set t, t Z. Thus the positive solutions which lie below 5000 are 74, 1064, 2054, 3044, Example 5.7. Find the smallest positive integer leaving remainder 4, 5, 6 on division by 5, 6, 7 respectively. Solution: This is similar to the previous example. However, this time we don t have to work so hard. There is an obvious solution to the system of congruences, namely, c = 1. Thus the general solution is t (since = 210). Therefore the smallest positive solution is = 209.