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3240 MATH - INTRODUCTION TO NUMBER THEORY : HOMEWORK 7 Chapter 15. Unique Factorization A. Division Theorem. Problem 1 (Problem E1(i)). Find the quotient and reminder in Q[x]: Proof. x3 7x 1 = (x 2)(x2 + 2x 3) + ( 7) so the quotient is x2 + 2x 3 and the remainder is 7. Notice that 7 = 23 7 2 1, by the remainder theorem. Problem 2 (Problem E4(ii)). Find the remainder when x4 7x2 + 3 is divided by x + 1 in Q[x]. Proof. By the Remainder theorem, the remainder is equal to ( 1)4 7 ( 1)2 + 3 = 1 7 + 3 = 3. Problem 3 (Problem E8). For which values of k in Q does x k divide f (x) = x3 kx2 2x + k + 3? Proof. By the root theorem, x k divides f (x) if and only if f (k) = 0. And: f (k) = k 3 k 3 2k + k + 3 = k + 3 so f (k) = 0 if and only if k = 3. So k = 3 is the only value. C. GCD S. Problem 4 (Problem E1). Using Euclid s algorithm, nd a greatest common divisor in F3 [x] of: Proof. (1) x5 + 1 = (x2 + 1) (x3 x) + (x + 1) x2 + 1 = (x + 1) (x 1) + 2. so the GCD is 2 mod 3 (i.e. a unit). (2) Notice that x3 + 2x2 + 3x + 2 x3 + 2x2 + 2. Thus: x3 + 2x2 + 2 (x2 x + 4)(x + 3) + ( x 10) (x2 + 2x + 1)x + 2x + 2 x2 + 2x + 1 ( x 1)( x 1) + 0. Therefore, a GCD is 2x + 2 mod 3 (or x + 1 mod 3). Problem 5 (Problem E6(i)). In Z/3Z[x], write if possible the polynomial 1 in the form f (x)p(x) + g(x)q(x) = 1 where 1 2 MATH 3240 - INTRODUCTION TO NUMBER THEORY : HOMEWORK 7 Proof. p(x) = x3 + 1, First, we use Euclid s algorithm: q(x) = x3 + x + 1. x3 + x + 1 = (x3 + 1) 1 + x x3 + 1 = x x2 + 1. Now work backwards: 1 = x3 + 1 x x2 = x3 + 1 (x3 + x + 1 (x3 + 1)) x2 = (1 + x2 )(x3 + 1) + ( x2 )(x3 + x + 1). Thus, f (x) = 1 + x2 and g(x) = x2 . D. Factorization into irreducible polynomials. Problem 6 (Problem E1). Any polynomial of degree 1 in F [x], where F is a eld, is either irreducible or it factors into a product of irreducibles. Proof. We prove this using induction on n = deg(f (x)). Suppose n = deg(f ) = 1. Then f (x) is an irreducible polynomial because if f (x) = p(x)q(x) then 1 = deg(f ) = deg(p) + deg(g). Note that the previous equality is only true because F is a eld. Therefore, deg(p) = 1 and deg(q) = 0, or deg(p) = 0 and deg(q) = 1. Hence, either p or q is a unit polynomial (if deg(p(x)) = 0 then p(x) is a non-zero constant). Now, suppose the result is true for all polynomials g(x) F [x] of degree 1 deg(g) n, and suppose f (x) is a polynomial of degree n + 1. If f (x) is an irreducible, then we are done. Otherwise, f (x) is reducible, i.e. f (x) = p(x)q(x) with 1 p(x), q(x) n and deg(p(x)) + deg(q(x)) = deg(f (x)) = n + 1. By the induction hypothesis, both p(x) and q(x) are irreducibles or the product of irreducibles: p(x) = r1 (x) r2 (x) rt (x), q(x) = s1 (x) s2 (x) sm (x) with ri (x) and sj (x) irreducible polynomials in F [x]. Hence: f (x) = p(x)q(x) = r1 (x) r2 (x) rt (x) s1 (x) s2 (x) sm (x) and therefore f (x) is a product of irreducibles. Problem 7 (Problem E9). Factor x5 x into a product of irreducibles in Z/5Z[x]. Proof. Let f (x) = x5 x. Recall that by the root theorem, if f (a mod 5) 0 mod 5 then (x a) divides f (x) in Z/5Z[x]. Moreover, by Fermat s little theorem, we know that a5 a mod 5, for all a 0, 1, 2, 3, 4 mod 5. Therefore, a 0, 1, 2, 3, 4 mod 5 are all roots of x5 x and, hence, (x a) divides x5 x for a = 0, 1, 2, 3, 4, in Z/5Z[x]. Since (x 0)(x 1)(x 2)(x 3)(x 4) is a monic polynomial of degree 5 that divides x5 x, they must be equal. Hence: x5 x x(x 1)(x 2)(x 3)(x 4) mod 5. Problem 8 (Problem E10). Show that for any prime p the polynomial xp x factors as x(x 1)(x 2) (x (p 1)) over Z/pZ[x]. Proof. Let f (x) = xp x. Recall that by the root theorem, if f (a mod p) 0 mod p then (x a) divides f (x) in Z/pZ[x]. Moreover, by Fermat s little theorem, we know that a5 a mod p, for all a 0, 1, 2, . . . , p 1 mod p. Therefore, a 0, 1, 2, . . . , p 1 mod p are all roots of xp x and, hence, (x a) divides xp x for a = 0, 1, 2, . . . , p 1, in Z/pZ[x]. Since MATH 3240 - INTRODUCTION TO NUMBER THEORY : HOMEWORK 7 3 (x 0)(x 1)(x 2) (x (p 1)) is a monic polynomial of degree p that divides xp they x, must be equal. Hence: xp x x(x 1)(x 2) (x (p 1)) mod p. Chapter 23. Primitive Roots A. Primitive roots modulo m. Problem 9 (Problem E3). Prove that 74 is a primitive root modulo 89. Proof. First we show that 2 has order 11 modulo 89. Notice that if we show that 211 1 mod 89, then the order must be 11 because the order would divide 11 and it is clearly not just 1, so it must be 11. In order to show that 211 1 mod 89, notice that 26 64 25 (52 ) mod 89. Moreover 54 (252 ) 625 2 mod 89. Therefore: 212 (26 )2 ( 52 )2 54 2 mod 89 and so, 211 1 mod 89. Next we show that 37 has order 8 modulo 89. Calculate 372 34 mod 89 and 342 88 1 mod 89. Therefore 378 (374 )2 ( 1)2 1 mod 89. Finally, since ord(2) = 11, ord(37) = 8 and (11, 8) = 1, it follows that ord(74) = ord(2 37) = 11 8 = 88 = 89 1. Hence, 74 is a primitive root modulo 89. Problem 10 (Problem E4). Find a primitive root modulo 61. Proof. Let us check that 2 is a primitive root modulo 61. Thus, we need to check that the order of 2 is exactly 60. Notice that the order of 2 must be a divisor of 60 = 4 3 5, so the possible orders are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. We need to check that 2d = 1 mod 61 for all d = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 but 260 1 mod 61 (the last congruence is, of course, a result of Fermat s little theorem and it doesn t need to be checked). 2 22 23 24 25 26 210 212 215 220 230 260 = 1 mod 61, 4 = 1 mod 61, 8 = 1 mod 61, 16 = 1 mod 61, 32 = 1 mod 61, 64 3 = 1 mod 61, 26 24 3 16 48 = 1 mod 61, 210 22 48 4 192 9 = 1 mod 61, 212 23 9 8 11 = 1 mod 61, 215 25 11 32 352 47 = 1 mod 61, (215 )2 112 121 1 = 1 mod 61, (230 )2 ( 1)2 1 mod 61. 4 MATH 3240 - INTRODUCTION TO NUMBER THEORY : HOMEWORK 7 Problem 11 (Problem E7). Find a primitive root modulo 73. Proof. We begin by calculating the order of 2 modulo 73. Notice that the possible orders are the divisors of 72 = 23 32 , which are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. After some calculations, we nd that 29 1 mod 73 and not before. Thus, the order of 2 is 9, not a primitive root. Let us try 3 next. After the appropriate calculations, we nd that 312 1 mod 73 and not before. Therefore the order is 12. Since (12, 9) = 3, we use 3 to nd another congruence of order 4. Since 3 has order 12 then 33 = 27 must have order 4. Now, if we had instead an element a of order 8, then we would be almost done because 2a would have order 8 9 = 72. Since 27 has order 4, if we have a such that a2 27 then a would have order 8. So we try to nd a root of x2 27 mod 73. It turns out that 102 27 mod 73. And we can check that the order of 10 is precisely 8 modulo 73. Since 8 and 9 are relatively prime, and ord(2) = 9, ord(10) = 8, it turns out that ord(20) = ord(2 10) = 8 9 = 72, by a result in class. Therefore, 20 is a primitive root modulo 73. Problem 12 (Problem E9). Let p be an odd prime. Show that if b is a primitive root modulo p then b(p 1)/2 1 mod p. Proof. Let p be an odd prime, let b be a primitive root modulo p, notice that (p 1)/2 is an integer (because p is odd) and put a b(p 1)/2 First, we claim that a2 1 mod p. Indeed: a2 (b(p 1)/2 )2 bp 1 1 mod p by Fermat s little theorem. However, we know that x2 1 mod p has only two solutions, namely 1. But since b is a primitive root, we cannot have b(p 1)/2 1 mod p because this would contradict the fact that the order of b is precisely p 1. Therefore, a b(p 1)/2 1 mod p as claimed. Problem 13 (Problem E10). Prove Wilson s theorem using primitive roots. Proof. Let p be an odd prime and let b be a primitive root modulo p. Then the order of b is precisely p 1 and, therefore, every unit 1, 2, . . . , p 1 modulo p can be expressed as one of the powers: b, b2 , b3 , . . . , bp 1 mod p. Therefore, {1, 2, . . . , p 1} and {b, b2 , . . . , bp 1 } are both complete systems of representatives of the units modulo p and so: (p 1)! 1 2 (p 1) b b2 bp 1 b1+2+ +(p 1) bp(p 1)/2 (b(p 1)/2 )p ( 1)p 1 modulo p, where we have used Problem 9 and the equality 1 + 2 + 3 + . . . + n = n(n+1) . 2 mod p.

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