1 MATH 310: Homework 7 Due Thursday, 12/1 in class Reading: Davenport III.1, III.2, III.3, III.4, III.5 1. Show that x is a root of unity modulo m if and only if (x, m 1. (Hint: Use Euler s theorem and #4 of HW5 To say that x is a root of unity means x n 1 for some n > 0. Hence, x n 1 + km for some k, which implies (x n, m 1. Using #4 of HW5, we conclude (x, m 1. Conversely, if (x, m 1 then by Euler s Theorem x φ(m 1 (mod m so that x is a root of unity. 2. Find all primitive roots mod 11. Using #1 and #2 of HW8 with p 11: 2 5 1 implies 2 is primitive; 3 5 (3 2 (3 3 ( 2( 6 1 implies 3 is not; 4 5 (2 5 2 1 implies 4 is not; 5 5 (5 2 (5 3 (3(4 1 implies 5 is not; 6 5 (2 5 (3 5 1 implies 6 is primitive; 7 5 ( 4 5 1 implies 7 is primitive; 8 5 ( 3 5 1 implies 8 is primitive; 9 5 ( 2 5 1 implies 9 is not; 10 2 1 implies 10 is not. Total # should be φ(φ(11 4: 2,6,7,8 3. Find all primitive roots mod 27. There are φ(φ(27 φ(18 6 of these. Using the results from the section More Bonus Stuff of HW8 solutions, we only need to check that g 9 1 and g 3 1. Using this, we see that 2 is a primitive root since 2 3 8 and 2 9 1. We can skip checking any g such that 3 g, since g 3 0 (mod 27. We also know that 4 cannot be primitive, since it is a square. We see that 5 is a primitive since 5 3 10 and 5 9 1. Next, 7 3 8 and 7 9 1 shows that 7 is not primitive. However, this calculation shows that 7 20 must therefore be primitive. Note also we can see from this calculation that ±8 8, 19 cannot be primitive (because they are cubes. Similarly, ±10 10, 17 cannot be primitive.
2 Similarly, we find 11 is primitive because 11 3 8 and 11 9 1, while 13 is not because 13 3 10 and 13 9 1, but that means 13 14 is primitive. Only one possibility remains, namely 23 4, which is now easily seen to be primitive. Hence, the 6 primitive roots to the modulus 27 are: 2, 5, 11, 14, 20, 23. 4. Find a primitive root to the modulus 73. (Hint: 10 and 2 are solutions to x 8 1 and x 9 1 (mod 73, respectively. Note that φ(73 72 8 9. Recall that the order of the product of two numbers to any modulus is the product of their orders provided they are relatively prime. (This was used in the proof of the multiplicative property of the Euler φ-function. Hence, it is enough to show that 10 has order 8 and 2 has order 9. To show that 10 has order 8, it is enough to verify that 10 4 1 (mod 73. Indeed, 10 2 100 27 so that 10 4 729 1 1. Similarly, to see that 2 has order 9, we check that 2 3 1 (mod 73, which is clear. 5. Show that if g is a primitive root to the modulus p 2, then it is also a primitive root to the modulus p. Is the converse true? Let g be a primitive root to the modulus p 2. This is equivalent to saying g r 1 (mod p 2 for any 0 < r < φ(p 2 p(p 1. Proceeding by contradiction, suppose g is not a primitive root to the modulo p. That means there exists 0 < h < p 1 such that g h 1 (mod p. Hence, g h 1 + kp for some integer k. Raising both sides to the power p, we obtain g ph (1 + kp p 1 + p 2 (k +... 1 (mod p 2. But this is a contradiction, since 0 < ph < φ(p 2. The converse holds for p 3 because 2 is the only primitive root to the modulus 3 and it happens to be primitive to modulus 9 since it has φ(9 6 distinct powers: 2, 4, 8, 7, 5, 1 The converse also holds for p 5 because the two primitive roots to the modulus 5 are ±2 which happen to be primitive to modulus 25 since each has φ(25 20 distinct powers: ±2, 4, ±8, 9, ±7, 11, ±3, 6, ±12, 1, 2, 4, 8, 9,..., 1.
Alternatively, we can use More Bonus Stuff in HW8 solutions and simply check that (±2 5 (±7 so that (±2 10 (±7 2 1, which implies both are primitive to the modulus 25. When p 7, we get a counterexample to the converse using either primitive root g 3 or 5 to the modulus 7, neither of which is primitive to modulus 49 since 3 3 22, 3 4 17, 3 7 18 so that 3 21 396 4 1 (mod 49 and 5 3 27, 5 4 12, 5 7 (54( 6 30 19 so that 5 ( 21 (9(8 1 (mod 49. 6. Find all sets of two decimal digits which can occur as the last two digits of a perfect square. Let n be a square. The problem asks for the possible values of n (mod 100. In other words, we are asked to find all squares to the modulus 100. Since n is a square to the modulus 100, it is also a square to the modulus 4. Hence, n (mod 4 0, 1. Similarly, n (mod 5 0, 1, 4. From this, it follows according to the 3 possibilities that n (mod 25 0 or 1, 6, 11, 16, 21 or 4, 9, 14, 19, 24. By Chinese Remainder Theorem, n (mod 100 0, 25 or o6, e1 or e4, e9 (where e and o stand for any even or odd digit, respectively giving a total of 22 squares to the modulus 100. ( ( ( 26 19 33 7. Calculate the Legendre symbols, and. ( 26 73 ( ( 33 3 73 73 73 73 73 ( ( ( ( ( 1 2 13 73 8 73 73 73 13 13 ( ( ( ( 2 19 73 16 4 1 73 19 19 19 ( ( ( ( 11 73 7 11 73 11 11 7 ( 2 13 1 3 ( 4 1 7 8. Determine which of the following congruences are soluble: (a x 2 125 (mod 1016 Since 1016 2 3 127, the congruence has a solution if and only if it has a solution to the modulus 8 as well as to the modulus 127. Since ( ( ( ( 125 2 1 2 ( 1(+1 1 127 127 127 127
4 it follows that 125 is not a square to the modulus 127 (in other words, not a quadratic residue. Hence, (a is insoluble. (b x 2 129 (mod 1016 Since 129 1 (mod 8, 129 is a square to the modulus 8. That is, there exists a residue a to the modulus 8 such that a 2 129. (It s obvious that a ±1, but the purpose ( of introducing ( the a will become clear 129 2 shortly. Similarly, since 1, 129 is also a square 127 127 to the modulus 127. Let b be a residue to the modulus 127 such that b 2 129 (mod 127. (Here, it is not as obvious how to determine b, but doing that is not the point. By the Chinese Remainder Theorem, there is a residue c to the modulus 1016 such that c a (mod 8 and c b (mod 127. (The fact that c is uniquely determined is irrelevant for this problem. Moreover, c 2 129 (mod m holds for m 8 and m 127. Since (8, 127 1, it follows that (b is soluble. (c x 2 41 (mod This is insoluble since ( ( ( 41 3 41 41 ( ( 1 3 41 41 ( 41 3 ( 2 1 3 (d 41x 2 43 (mod Method 1. First, observe that the quadratic character of a residue, i.e. whether or not it is a square, is the same as that of its multiplicative inverse. In other words, if xy 1 (mod p then x is a square to the modulus p if and only if y is a square to the modulus p. The quickest way to see this is to use the fact ( xy p ( x p ( y p to conclude that the RHS of this identity is one. Alternatively, one can argue that the sum of the indices of x and y (relative to any primitive root of unity vanishes so that one is even if and only if the other is. 1 Second, determine the quadratic character of 43 by the calculation ( ( ( ( ( ( 43 1 7 43 1 1 43 43 43 7 7 1 The latter approach uses the existence of primitive roots of unity, which we also used to establish the homomorphism property of the Legendre symbol.
5 or more quickly via ( 43 ( 36 ( 1 1. Now, let k be the solution to 41k 1 (mod. (In other words, let k be the inverse of 41 in the group (Z/Z. Note that the fact that k is exists (and is uniquely determined is a consequence of the fact that gcd(41, 1. Then (d is soluble if and only if x 2 43k (mod is soluble, which it is because by the previous calculation and the fact that (c is insoluble, we have ( 43k so that (d is soluble. ( 43 ( k ( 1( 1 1 Method 2. First, multiply (d by 2 to get an equivalent congruence 3x 2 7 (mod, then multiply by -26 to get x 2 182 (mod, then calculate ( ( ( ( ( ( 182 55 5 11 4 2 1. 5 11 (e 43x 2 47 (mod Follow Method 1 and compute ( ( 47 32 to conclude that (e is soluble. ( 1 ( 2 ( 1(+1 1 Or, follow Method 2 and first multiply by 2 to get 7x 2 15 (mod, then multiply by 11 to get 2x 2 7 (mod, then multiply by 39 to get x 2 273 (mod, then compute ( 273 ( 36 1 (f x 2 151 (mod 840 Since 8 divides 840 and x 2 1 (mod 8 is insoluble, (f is insoluble.
6 9. (BONUS Show that a primitive root of unity modulo a prime p cannot be a quadratic residue modulo p. Let g be a primitive root of unity modulo p. For any x 0 (mod p, x is a quadratic residue modulo p if and only if its index relative to g is even. Since g has index one relative to itself, it is not a quadratic residue.