M381 Number Theory 2004 Page 1

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1 M81 Number Theory 2004 Page 1 [[ Comments are written like this. Please send me (dave@wildd.freeserve.co.uk) details of any errors you find or suggestions for improvements. ]] Question 1 20 = 2 * = 4 * = * = 2 * + 1 = * Therefore as gcd(20, 10) = 1 then 20 and 10 are relatively prime. 1 = 7 2 * = 7-2 * (24 - * 7) = 7 * 7-2 * 24 = 7 * (10-4 * 24) 2 * 24 = 7 * 10-0 * 24 = 7 * 10-0 * (20-2 * 10) = 67 * 10-0 * 20. Therefore 10x - 20y = 1 when x = 67 and y = 0. Let P(n) be the proposition ½ n(n -1) = ½ n 2 (n + 1). P(1) is 1 = ½ * 1 2 * (1 + 1). As P(1) is true then we have the basis for induction. Assume P(k) is true for some positive integer k ½ k(k -1) + ½ (k + 1)[(k + 1) -1] = ½ k 2 (k + 1) + ½ (k + 1)(k + 2) ( using the induction hypothesis) = ½ (k + 1) (k 2 + k +2) = ½ (k + 1) 2 (k +2). Therefore if P(k) is true then P(k + 1) is true. This completes the induction step. The result then follows from the Principle of Mathematical Induction. ( marks) gcd(a, b) = 1 gcd(a, -b) = 1 gcd(a, a - b) = 1. Therefore if c a -b then gcd(a, c) = 1. [[ Alternatively. Since gcd(a,b) = 1 then we can write 1 = a - b = ( - )a + (a - b), where and are integers. As c a - b then we can write 1 = ma + nc for some integers m and n. As 1 can be expressed in this form then gcd (a, c) = 1. ]]

2 M81 Number Theory 2004 Page 2 Question 2 (11 marks) False. 87 = 29 * = 17 * The prime divisors 29 and are not of the form 17m + 2. As 2 is of the form 17m + 2 then we only need to consider odd values of m. Any prime greater than can be expressed in the form 12k + 1, 12k + 5, 12k + 7, 12k (mod 12). Therefore if p > is a prime then p 2 is of the form 12k + 1. Therefore the statement is true. Any twin primes greater than must either be of the form 12m + 5, 12m + 7 or 12m + 11, 12(m + 1) + 1. In both cases the sum of the primes is divisible by 12. Therefore the statement is true.

3 M81 Number Theory 2004 Page Question (1 mark) Since n 5 (mod 10) then n = 10k + 5 for some integer k. Therefore 6n + 1 = 6(10k + 5) + 1 = 60k + 1. As 6n (mod 12) then the least positive residue modulo 12 of 6n + 1 is 7. Let 0 i, j n -1. If c + ia c + ja (mod n) then ia ja (mod n). Since gcd(a, n) = 1 then a can be cancelled to give i j (mod n). Therefore none of the values c, c + a, c + 2a,..., c + (n - 1)a are congruent modulo n. As there are n of these values then they form a complete set of residues modulo n. (6 marks) By the Chinese Remainder theorem these equations have a unique solution modulo *4*1 = 156. Integers which satisfy the congruence x 7 (mod 1) are 7, 20,,... Integers which also satisfy the congruence x 1 (mod 4) are, 85, 17,... Integers which also satisfy the congruence x 2 (mod ) are 17,... Therefore the least positive integer which satisfies the linear congruences is 17.

4 M81 Number Theory 2004 Page 4 Question 4 ( marks) By FLT, (mod 7). Therefore (10 6 ) 2 * * 100 * * (mod 7). Therefore the least positive residue is 1. (a) 12x 1 (mod 7) 6x (mod 7) -x (mod 7) x 4 (mod 7) By FLT, (mod 7). Therefore as 12 * (mod 7) then 12 5 (mod 7) is also a solution of 12x 1 (mod 7). As this equation has a unique solution then 4 is equal to the least positive residue of 12 5 (mod 7). (b) 12 2 x 144x x 1 (mod 7) has a solution congruent to 12 4 (mod 7) x 1 (mod 7) (6) 2 x 2 (mod 7) (-1) 2 x 9 (mod 7) x 9 (mod 7) Therefore 9 is equal to the least positive residue of 12 4 (mod 7).

5 M81 Number Theory 2004 Page 5 Question 5 (5 marks) If 2 p - 1 is prime Since 2 p - 1 is prime then gcd(2 p - 1, 2 p - 1 ) = 1 and σ(2 p - 1) = p - 1 = 2 p. σ(2 p 1 (2 p - 1)) = σ(2 p - 1 ) σ(2 p - 1) as σ multiplicative = (2 p - 1) 2 p = 2 {2 p - 1 (2 p - 1)} A number n is perfect if σ(n) = 2n. Therefore 2 p - 1 (2 p - 1) is perfect if 2 p - 1 is prime. If 2 p - 1 (2 p - 1) is perfect As 2 p - 1 is odd then gcd (2 p - 1, 2 p - 1) = 1. σ(2 p - 1 (2 p - 1)) = σ(2 p - 1 ) σ(2 p - 1) Since σ is a multiplicative function = (2 p - 1) σ(2 p - 1) = 2 p (2 p - 1) as 2 p - 1 (2 p - 1) is perfect This means that σ (2 p - 1) = 2 p. As 1 and 2 p - 1 are factors of 2 p - 1 then σ(2 p - 1) 1 + (2 p - 1) = 2 p. As the equality only holds when 2 p - 1 is prime then 2 p - 1 is prime. Therefore 2 p - 1 (2 p - 1) is perfect if and only if 2 p - 1 is prime. (a) (2 marks) 110 = 11 * 5 * 2. σ(110) = σ(11) σ(5) σ(2) = 12 * 6 * = 72 * = = 4 * 28 = 2 4 * 7. σ(112) = σ(2 4 ) σ(7) = (2 5-1) * 8 = 1 * 8 = 248. Therefore only 112 is abundant. (b) If n = 4p 2 where p is an odd prime then σ(n) = σ(2 2 ) σ(p 2 ) = (2-1) (1 + p + p 2 ). If 4p 2 is abundant then 7(1 + p + p 2 ) > 8p 2. So p 2 < 7(p + 1) or Therefore 4p 2 is abundant only for the odd primes, 5, or 7. 7 p < 7 +. p

6 M81 Number Theory 2004 Page 6 Question 6 ( marks) The quadratic congruence has solutions if * 2 * 4 = 25-2 = -7 is a quadratic residue of 1. (-7/1) = (24/1) Th. 2.1(a) (mod 1) = (2 2 /1)(2/1)(/1) Th. 2.1(c) = 1 * 1 * (-1) Th. 2.1(b), Th..2 and Th = -1 Therefore the congruence does not have any solutions. (62/19) = (2/19)(1/19) Th. 2.1(c) = (-1) * -(19/1) Th..2 and LQR (mod 4) = (15/1) Th. 2.1(a) (mod 1) = (-16/1) Th. 2.1(a) (mod 1) = (-1/1) (4 2 /1) Th. 2.1(c). = -1 * 1 Th. 2.1(e) and Th. 2.1(b) = -1 (/p) = (p/) if p 1 (mod 4). LQR = - (p/) if p (mod 4). LQR. (p/) = (1/) = 1 p 1 (mod ). = (2/) = -1 p 2 (mod ). By the Division Algorithm an odd prime greater than must be of the form 12k +1, 12k + 5, 12k + 7, 12k +11. p (mod 12) p (mod 4) p (mod ) (/p) = (p/) = (1/) = = (p/) = (2/) = = - (p/) = - (1/) = = - (p/) = - (2/) = (-1) 2 = 1 Therefore the given result holds.

7 M81 Number Theory 2004 Page 7 Question 7 82 = 1 * = 5 * = * = 4 * Therefore 82/69 = [1, 5,, 4] = [1, 5,,, 1]. (7 marks) Let α = [0, 2, x] where x = [<2, 1>] = [2, 1, x]. The convergents of [2, 1, x] are 2/1, /1, (x + 2)/(x + 1) = x So x 2-2x - 2 = 0 and this has the positive solution x = = The convergents of [0, 2, x] are 0/1, 1/2, x/(2x + 1) = α. 1 + This gives [ ] ( 1 + )( 2 ) ( 6) + ( 2) 0,2, 2,1 = = = = α = [0, 2, 2, 1, 2, 1, 2, 1, <1, 2>]. Convergents of α are C 1 = 0/1; C 2 = 1/2; C = 2/5; C 4 = /7; C 5 = 8/19; C 6 = 11/26; C 7 = 0/71. By Corollary to Theorem q q k k + 1 p k < α <, where C k = p k / q k. q k 1 q q k k + 1 When k = 5 we have α - C 5 < 1/(19 * 26) < 1/400. When k = 4 we have 1/(2*7*19) = 1/(14 * 19 ) =1/266 < α - C 4. Therefore the 5th convergent 8/19 is the 1st convergent within 1/400 of [0,2, <2, 1> ]

8 M81 Number Theory 2004 Page 8 Question 8 Since the cycle length of 11 is 2 then every even convergent gives a solution of the Diophantine equation (Th. 1.2). Convergents of [, <, 6>] are C 1 = /1; C 2 = 10/; C = 6/19; C 4 = 199/60. Therefore 2 positive solutions are x = 10, y = ; and x = 199, y = = (200-1) 2 = 40, = 9, * 60 2 = 11 * 600 = 9,600. A primitive Pythagorean triple is of the form (2mn, m 2 n 2, m 2 + n 2 ), where m and n are positive integers, m > n, gcd(m, n) = 1, and m and n have opposite parity (Th. 2.1). As the 2nd and rd sides are odd then we must have 2mn = 12. As mn = 6 then m = 6, n = 1, and m =, n = 2 are the only possibilities. Therefore the only primitive Pythagorean triples are (12, 5, 7) and (12, 5, 1). Since (4,, 5) is a primitive Pythagorean triple then (16, 12, 20) is a non-primitive Pythagorean triple with 12 as a side. ( marks) Assume that x = x 1, y = y 1, z = z 1 is a solution in positive integers. Therefore x 1 + 9y 1 = z 1 Since the other 2 terms in the equation are divisible by then x 1 is also divisible by. Therefore we can write x 1 = x 2 where x 2 is a positive integer. Therefore 27x 2 + 9y 1 = z 1. Dividing by gives 9x 2 + y 1 = z 1. Similarly z 1 is also divisible by. Therefore we can write z 1 = z 2 where z 2 is a positive integer. Hence 9x 2 + y 1 = 27z 2. Dividing by gives x 2 + y 1 = 9z 2. Similarly y 1 is also divisible by. Therefore we can write y 1 = y 2 where y 2 is a positive integer. So x y 2 = 9z 2. Dividing by gives x 2 + 9y 2 = z 2. Therefore x = x 2, y = y 2, z = z 2 is also a solution of x + 9y = z with z 2 < z 1 in positive integers. As the descent step has been established then by the method of infinite descent there can be no solution in positive integers. END OF PART 1 SOLUTIONS

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