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Your Name: MA 261 — Worksheet Monday, April 21, 2014 1. Show that 211,213 − 1 is not divisible by 11. Solution: To say that 211,213 − 1 is divisible by 11 is equivalent to saying that 211,213 ≡ 1 (mod 11). Thus, to show our statement is equivalent to showing that 211,213 is congruent to something different from 1 modulo 11. Let us take (small) consecutive powers of 2 and see if there is a pattern that we can use to simplify our calculations. Note that 20 ≡ 1 (mod 11) 21 ≡ 2 (mod 11) 22 ≡ 4 (mod 11) 23 ≡ 8 (mod 11) 24 ≡ 5 (mod 11) 25 ≡ 10 (mod 11) 26 ≡ 9 (mod 11) 27 ≡ 7 (mod 11) 28 ≡ 3 (mod 11) 29 ≡ 6 (mod 11) 210 ≡ 1 (mod 11). Observe that we could have simplified are calculation as follows: 25 ≡ 10 (mod 11) ; 25 ≡ −1 (mod 11) ; 210 = (25 )2 ≡ (−1)2 = 1 (mod 11). No matter what, we have shown that 2 raised to the tenth power is congruent to 1 modulo 11. If we divide the exponent 11, 213 by 10 we obtain 11, 213 = 10 · 1, 121 + 3. Thus 211,213 = 210·1,121+3 = (210 )1,121 · 23 ≡ (1)1,121 · 23 ≡ 1 · 8 ≡ 8 (mod 11). Thus, we have shown that 211,213 ≡ 8 (mod 11), which implies that 211,213 − 1 cannot be divisible by 11. The above calculations shows that 211,213 − 8 is divisible by 11. 2. (1) Describe all solutions of the congruence 22x ≡ 5 (mod 15). Solution 1 (classic method): By Theorem 3.19 this congruence has a solution if and only if there exist integers x and y such that 22x + 15y = 5. We know that this Diophantine equation has a solution if and only if gcd(22, 15) divides 5, by Theorem 1.48 (see also Theorem 3.20). As gcd(22, 15) = 1, we do have a solution. Moreover, all the solutions are described by Theorem 1.53. Namely, x = x0 + 15k gcd(22, 15) y = y0 − 22k gcd(22, 15) k ∈ Z, where x0 and y0 are a particular solution of 22x + 15y = 5. Do not memorize those formulas! For your instruction, we repeat next the proof in this specific example. To find a particular solution we use the Euclidean Algorithm (Exercise 1.35) applied to a = 22 and b = 15. We have: 22 = 15 · 1 + 7 15 = 7 · 2 + 1 1 7 = 7 · 1 + 0. 2 This says that the last nonzero remainder is 1; this is the gcd between 22 and 15. We can also use the above equations to express 1, the gcd, as a linear combination of 22 and 15, as follows: 1 = 15 − 7 · 2 ; 1 = 15 − (22 − 15) · 2 ; 1 = 15 · 3 + 22 · (−2). If we multiply the last equation by 5 we obtain 22 · (−10) + 15 · (15) = 5. (1) Thus x0 = −10 and y0 = 15. Consider now, another set of integer solutions x and y of 22x + 15y = 5. If we subtract equation (1) from 22x + 15y = 5 we obtain 22 · (x + 10) + 15 · (y − 15) = 0 ⇐⇒ 15 · (15 − y) = 22 · (x + 10). (2) This means that 15 divides the product 22 · (x + 10). However, by Theorem 1.41 we have that 15 divides x + 10, as gcd(15, 22) = 1. Thus x + 10 = 15 · k ⇐⇒ x = −10 + 15 · k, k ∈ Z. Substituting back in equation (2) we obtain that y = 15 − 22 · k, even if we don’t need the values for y. Thus the solutions of 22x ≡ 5 (mod 15) are x = −10 + 15 · k where k is any integer. The solution set consists of the numbers {. . . , −25, −10, 5, 20, 35, . . . }. Observe, tough, that the only solution that satisfies 0 ≤ x ≤ 14 is x = 5 . This confirms the results in Theorem 3.24. There is exactly 1 = gcd(22, 15) solution in the range 0 ≤ x ≤ 14. Solution 2 (easier, but requires some tricks): Notice that, by reducing the coefficient 22 modulo 15, the original congruence is equivalent to 22x ≡ 5 (mod 15) ⇐⇒ 7x ≡ 5 (mod 15). Now, multiply both sides of the latter congruence by 2. We obtain 14x ≡ 10 (mod 15) ⇐⇒ −x ≡ 10 (mod 15), as 14 ≡ −1 (mod 15). Multiply both sides by −1 and observe that −10 ≡ 5 (mod 15): x ≡ −10 (mod 15) ⇐⇒ x≡5 (mod 15). Hence our solutions are of the form x = 5 + 15ℓ, where ℓ ∈ Z. (2) Describe all solutions of the congruence 45x ≡ 15 (mod 24). Solution 1 (classic method): By Theorem 3.19 this congruence has a solution if and only if there exist integers x and y such that 45x + 24y = 15. We know that this Diophantine equation has a solution if and only if gcd(45, 24) divides 15, by Theorem 1.48 (see also Theorem 3.20). As gcd(45, 24) = 3, we do have a solution. Moreover, all the solutions are described by Theorem 1.53. Namely, x = x0 + 24k gcd(45, 24) y = y0 − 45k gcd(45, 24) where x0 and y0 are a particular solution of 45x + 24y = 15. k ∈ Z, 3 Do not memorize those formulas! For your instruction, we repeat next the proof in this specific example. To find a particular solution we use the Euclidean Algorithm (Exercise 1.35) applied to a = 45 and b = 24. We have: 45 = 24 · 1 + 21 24 = 21 · 1 + 3 21 = 3 · 7 + 0. This says that the last nonzero remainder is 3; this is the gcd between 45 and 24. We can also use the above equations to express 3, the gcd, as a linear combination of 45 and 24, as follows: 3 = 24 − 21 ; 3 = 24 − (45 − 24) ; 3 = 24 · 2 + 45 · (−1). If we multiply the last equation by 5 we obtain 45 · (−5) + 24 · (10) = 15. (3) Thus x0 = −5 and y0 = 10. Consider now, another set of integer solutions x and y of 45x + 24y = 15. If we subtract equation (3) from 45x + 24y = 15 we obtain 45 · (x + 5) + 24 · (y − 10) = 0 ⇐⇒ 8 · (10 − y) = 15 · (x + 5), (4) after we divided both sides by 3. This means that 8 divides the product 15·(x+5). However, by Theorem 1.41 we have that 8 divides x + 5, as gcd(8, 15) = 1. Thus x+5=8·k ⇐⇒ x = −5 + 8 · k, k ∈ Z. Substituting back in equation (4) we obtain that y = 10 − 8 · k, even if we don’t need the values for y. Thus the solutions of 45x ≡ 15 (mod 24) are x = −5 + 8 · k, where k is any integer. The solution set consists of the numbers {. . . , −21, −13, −5, 3, 11, 19, 27, . . . }. Observe, tough, that the only solutions that satisfies 0 ≤ x ≤ 23 are x = 3, 11, 19 . This confirms the results in Theorem 3.24. There are exactly 3 = gcd(45, 24) distinct solutions in the range 0 ≤ x ≤ 23. Solution 2 (easier, but requires even more tricks): Notice that, by reducing the coefficient 45 modulo 24, the original congruence is equivalent to 45x ≡ 15 (mod 24) ⇐⇒ −3x ≡ 15 (mod 24). Now, multiply both sides of the latter congruence by −1. We obtain 3x ≡ −15 (mod 24) ⇐⇒ 3x ≡ 9 (mod 24), as −15 ≡ 9 (mod 24). By Theorem 3.19, the congruence 3x ≡ 9 (mod 24) has a solution if and only if there exist integers x and y such that 3x + 24y = 9. All terms are divisibile by 3, so that x + 8y = 3. Hence, by using Theorem 3.19 again, we conclude that x is a solution of the simpler congruence x ≡ 3 (mod 8) 3x ≡ 9 (mod 24) ⇐⇒ x ≡ 3 (mod 8). Hence our solutions are of the form x = 3 + 8ℓ, where ℓ ∈ Z. The only solutions that satisfy 0 ≤ x ≤ 23 are again x = 3, 11, 19. 4 3. (1) Solve the simultaneous system of congruences x ≡ 1 (mod 8) x≡2 x ≡ 3 (mod 81). (mod 25) Solution: Notice that the numbers 8, 25, and 81 are pairwise relatively prime. Using the proof that we gave for the Chinese Remainder Theorem (see Theorem 3.29), we are looking for a solution of the form 8 · 25 · 81 8 · 25 · 81 8 · 25 · 81 x1 + x2 + x3 , x= 8 25 81 that is, x = 25 · 81 · x1 + 8 · 81 · x2 + 8 · 25 · x3 , (5) for suitable x1 , x2 , and x3 . If we now substitute the proposed solution described in (5) into the three given congruences we realize that we need to solve the following independent (from each other) congruences 25 · 81 · x1 ≡ 1 (mod 8) 8 · 81 · x2 ≡ 2 (mod 25) 8 · 25 · x3 ≡ 3 (mod 81). The first congruence reduces to 25 · 81 · x1 ≡ 1 (mod 8) ⇐⇒ 1 · 1 · x1 ≡ 1 (mod 8), because both 25 and 81 are congruent to 1 modulo 8. Thus we can choose x1 = 1 as a solution. The second congruence reduces to 8 · 81 · x2 ≡ 2 (mod 25) ⇐⇒ 8 · 6 · x2 ≡ 2 (mod 25), because 81 is congruent to 6 modulo 25. Again notice that we can simplify the congruence as follows 48 · x2 ≡ 2 (mod 25) ⇐⇒ −2 · x2 ≡ 2 (mod 25), as 48 is congruent to −2 modulo 25. If we multiply the latter congruence by −13 we obtain −2 · x2 ≡ 2 (mod 25) ⇐⇒ 26 · x2 ≡ −26 (mod 25). Since 26 ≡ 1 modulo 25 we have that x2 ≡ −1 (mod 25). Thus we can choose x2 = −1 as a solution. The third congruence reduces to 8 · 25 · x3 ≡ 3 (mod 81) ⇐⇒ 38 · x3 ≡ 3 (mod 81), because 8 · 25 = 200 is congruent to 38 modulo 81. If we multiply the latter congruence by 32 we obtain 32 · 38 · x3 ≡ 32 · 3 (mod 81) ⇐⇒ x3 ≡ 15 (mod 81). since 32 · 38 = 1216 ≡ 1 and 32 · 3 = 96 ≡ 15 modulo 81. Thus we can choose x3 = 15 as a solution. I am sure you will ask: how did you know that multiplying by 32 would have given such an easy value? It has to do with the fact that gcd(38, 81) = 1. Hence if you apply the Euclidean Algorithm to the pair 81 and 38, you will see that 1 = 38 · 32 + 81 · (−15). Thus 38 · 32 ≡ 1 (mod 81). 5 Summing up what we did so far, we have x = 25 · 81 · 1 + 8 · 81 · (−1) + 8 · 25 · 15 = 4, 377 satisfies the given congruences. By Theorem 3.29 any other solution x′ is such that x′ ≡ 4, 377 (mod 8 · 25 · 81) x′ ≡ 4, 377 (mod 16, 200). ⇐⇒ In other words, the only solution x of the three given congruences such that 0 ≤ x ≤ 16, 199 is x = 4, 377 . (2) Solve the simultaneous system of congruences y ≡ 5 (mod 8) y ≡ 12 y ≡ 47 (mod 25) (mod 81). Solution: Modify the previous calculations on your own. You will look for a solution of the form y = 25 · 81 · y1 + 8 · 81 · y2 + 8 · 25 · y3 , (6) for suitable y1 , y2 , and y3 . You will obtain that y2 = −6 y1 = 5 y3 = 43 do work. Thus y = 15, 437. Again by Theorem 3.29 any other solution y ′ is such that y ′ ≡ 15, 437 (mod 16, 200). In other words, the only solution y of the three given congruences such that 0 ≤ y ≤ 16, 199 is x = 15, 437 . 4. (Challenge): Show that for every integer n, the number n33 − n is divisible by 15. Sketch of the solution: Notice that 15 = 3 · 5 and gcd(3, 5) = 1. Thus by Theorem 1.42 (or Theorem 2.25) it is enough to show that n33 − n is divisible by 3 and by 5, respectively. In other words we need to show n33 ≡ n (mod 3) and n33 ≡ n (mod 5). The above relations will follow once we prove that for any n one has n3 ≡ n (mod 3) and n5 ≡ n (mod 5). (7) Indeed, n33 = (n3 )11 ≡ n11 = (n3 )3 · n2 ≡ n3 · n2 ≡ n · n2 = n3 ≡ n (mod 3), and n33 = (n5 )6 · n3 ≡ n6 · n3 = n9 = n5 · n4 ≡ n · n4 = n5 ≡ n (mod 5). Now, to prove the congruences in equation (7) we have to distinguish various cases. When we consider the congruences modulo 3 we have the following three cases n ≡ 0 (mod 3) n ≡ 1 (mod 3) n ≡ 2 (mod 3) that we need to analyze. When we consider the congruences modulo 5 we have the following five cases n ≡ 0 (mod 5) n ≡ 1 (mod 5) that we need to analyze. n≡2 (mod 5) n≡3 (mod 5) n ≡ 4 (mod 5) 6 Let me verify for you (7) in one of these cases so that you understand the ‘trick’. For instance, suppose that n ≡ 2 (mod 5). Then by Theorem 1.18 we have that n5 ≡ 25 (mod 5). But 25 = 32 which is congruent to 2 modulo 5. Hence n ≡ 2 (mod 5) ; n5 ≡ 2 (mod 5). Now, by the symmetric (Theorem 1.10) and the transitive (Theorem 1.11) properties of congruences, the two congruences above yield n5 ≡ n (mod 5). The remaining cases can be verified similarly.