Your Name: MA 261 Worksheet Monday, April 21, 2014

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1 Your Name: MA 261 Worksheet Monday, April 21, Show that 2 11,213 1 is not divisible by 11. Solution: To say that 2 11,213 1 is divisible by 11 is equivalent to saying that 2 11,213 1 (mod 11). Thus, to show our statement is equivalent to showing that 2 11,213 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 (mod 11) (mod 11) (mod 11) (mod 11) (mod 11) (mod 11) (mod 11) (mod 11) (mod 11) (mod 11) (mod 11). Observe that we could have simplified are calculation as follows: (mod 11) (mod 11) 2 10 = (2 5 ) 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, Thus 2 11,213 = ,121+3 = (2 10 ) 1, (1) 1, (mod 11). Thus, we have shown that 2 11,213 8 (mod 11), which implies that 2 11,213 1 cannot be divisible by 11. The above calculations shows that 2 11,213 8 is divisible by (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 Namely, x = x k gcd(22, 15) y = y 0 22k gcd(22, 15) where x 0 and y 0 are a particular solution of 22x + 15y = 5. k Z, 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 = = =

2 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 (22 15) 2 1 = ( 2). If we multiply the last equation by 5 we obtain Thus x 0 = 10 and y 0 = ( 10) + 15 (15) = 5. (1) 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 = k, k Z. Substituting back in equation (2) we obtain that y = k, even if we don t need the values for y. Thus the solutions of 22x 5 (mod 15) are x = 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 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 = l, 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 Namely, x = x k gcd(45, 24) y = y 0 45k gcd(45, 24) where x 0 and y 0 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 = = = 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 (45 24) 3 = ( 1). If we multiply the last equation by 5 we obtain Thus x 0 = 5 and y 0 = ( 5) + 24 (10) = 15. (3) 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 = 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 = 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 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. 3

4 4 3. (1) Solve the simultaneous system of congruences x 1 (mod 8) x 2 (mod 25) x 3 (mod 81). 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 that is, x = for suitable x 1, x 2, and x x x x 3, 81 x = x x x 3, (5) 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 x 1 1 (mod 8) 8 81 x 2 2 (mod 25) 8 25 x 3 3 (mod 81). The first congruence reduces to x 1 1 (mod 8) 1 1 x 1 1 (mod 8), because both 25 and 81 are congruent to 1 modulo 8. Thus we can choose x 1 = 1 as a solution. The second congruence reduces to 8 81 x 2 2 (mod 25) 8 6 x 2 2 (mod 25), because 81 is congruent to 6 modulo 25. Again notice that we can simplify the congruence as follows 48 x 2 2 (mod 25) 2 x 2 2 (mod 25), as 48 is congruent to 2 modulo 25. If we multiply the latter congruence by 13 we obtain 2 x 2 2 (mod 25) 26 x 2 26 (mod 25). Since 26 1 modulo 25 we have that x 2 1 (mod 25). Thus we can choose x 2 = 1 as a solution. The third congruence reduces to 8 25 x 3 3 (mod 81) 38 x 3 3 (mod 81), because 8 25 = 200 is congruent to 38 modulo 81. If we multiply the latter congruence by 32 we obtain x (mod 81) x 3 15 (mod 81). since = and 32 3 = modulo 81. Thus we can choose x 3 = 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 = ( 15). Thus (mod 81).

5 Summing up what we did so far, we have x = ( 1) = 4, 377 satisfies the given congruences. By Theorem 3.29 any other solution x is such that x 4, 377 (mod ) 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, (2) Solve the simultaneous system of congruences y 5 (mod 8) y 12 (mod 25) y 47 (mod 81). Solution: Modify the previous calculations on your own. You will look for a solution of the form y = y y y 3, (6) for suitable y 1, y 2, and y 3. You will obtain that y 1 = 5 y 2 = 6 y 3 = 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, (Challenge): Show that for every integer n, the number n 33 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 n 33 n is divisible by 3 and by 5, respectively. In other words we need to show n 33 n (mod 3) and n 33 n (mod 5). The above relations will follow once we prove that for any n one has Indeed, and n 3 n (mod 3) and n 5 n (mod 5). (7) n 33 = (n 3 ) 11 n 11 = (n 3 ) 3 n 2 n 3 n 2 n n 2 = n 3 n (mod 3), n 33 = (n 5 ) 6 n 3 n 6 n 3 = n 9 = n 5 n 4 n n 4 = n 5 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) n 2 (mod 5) n 3 (mod 5) n 4 (mod 5) that we need to analyze.

6 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 n (mod 5). But 2 5 = 32 which is congruent to 2 modulo 5. Hence n 2 (mod 5) n 5 2 (mod 5). Now, by the symmetric (Theorem 1.10) and the transitive (Theorem 1.11) properties of congruences, the two congruences above yield n 5 n (mod 5). The remaining cases can be verified similarly.