PMA225 Practice Exam questions and solutions Victor P. Snaith

Transcription

1 PMA225 Practice Exam questions and solutions 2005 Victor P. Snaith November 9, 2005

2 The duration of the PMA225 exam will be 2 HOURS. The rubric for the PMA225 exam will be: Answer any four questions. You are advised not to answer more than four; if you do, only your best four will be counted. THE FOLLOWING ARE A SELECTION OF TYPICAL PMA225 EXAM QUESTIONS, BEARING IN MIND THAT 2005 IS THE FIRST YEAR OF THE COURSE 1. (i) You publish (n, e) = (205, 9) in the RSA directory and receive the number 19. Decode it. 10 marks (ii) Solve the congruence 20x 2 + x (modulo 205). 7 marks (iii) Find the remainder when 397! is divided by marks Solution to question 1: (i) 205 = 41 5 in prime factors. To decode we want d such that 9d 1 (modulo LCM(40, 4) so d = 9 will do. Then (modulo 5) 19 9 ( 1) while (modulo 41) ( 8) ( 4) It suffices to find an integer in the sequence 6, 47, 88, 129,... which is congruence to 4 (modulo 5). So the decoding is 129. (ii) By the Chinese Remainder Theorem we must solve 20x 2 +x+4 x 1 (modulo 5) and 20x 2 + x (modulo 41). Hence x 1 (modulo 5). Also multiplying by 2, 40x 2 + 2x (modulo 41) or equivalently (x 4)(x + 2) x 2 2x 8 0 (modulo 41). Hence x 4 (modulo 41) or x 2 (modulo 41). The simultaneous solutions of this are [86] 205 and [121] 205. (iii) In prime factors 2005 = By Wilson s Theorem 400! 1 (modulo 401) so that 397! or 397! ( 3) ( 2) ( 1) 1 so 6 397! 1 (modulo 401). Since 6 67 = 402 we have 397! 67 (modulo 401). Also 397! 0 (modulo 5). Now look for a multiple of 5 in the sequence 67, 468, 869, 1270,... so that 397! 1270 (modulo 2005) by the CRT. 1

3 2. (i) State the Law of Quadratic Reciprocity. 1 mark (ii) Determine whether the congruence 24x 2 + 7x (modulo 419) has a solution. 12 marks (iii) Let p > 3 be a prime. Show that the congruence x 4 + 7x (modulo p) does not have a solution if and only if p 11 (modulo 12). Find all the solutions in the cases p = 13, 29, marks Solution to question 2: (i) For p, q > 2 distinct primes ( ( ) q p) p (p 1)(q 1) q = ( 1) 4. (ii) 419 is prime. b 2 4ac = = We can solve the quadratic if this is a square (modulo 419). However ( ) ( ) = = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = = ( 1)( )( ) = ( 1 7 ) 1 11 = ( 1) = 1. So a solution exists. (iii) (x 2 +3)(x 2 +4) = x 4 +7x (modulo p) so the congruence does not have a solution if and only if ( ) ( ) ( ) ( ) ( ) 3 p = 1 = 4 p. Now 3 p = 1 3 p p = ( 1) ( ) (p 1)/2 (p 1)(3 1) p 3 ( 1) 4 =( p 3) = 1 if and only if p 2 (modulo 3). Also ( ) ( ) 4 p = 1 p = 1 if and only if p 3 (modulo 4). The only simultaneous solutions to these congruences are p 11 (modulo 12). For the other values we have: p = 13, x ±6, ±3 (modulo 13), p = 29, x ±5 (modulo 29), p = 43, x ±13 (modulo 43). 2

4 3. A positive integer n is said to be (a) abundant (b) deficient (c) perfect if the sum of its proper divisors is (a) great than n (b) less than n (c) equal to n, respectively. (i) If p is a positive integer such that 2 p 1 is prime show that 2 p 1 (2 p 1)is perfect. 6 marks (ii) For each of 496, 945, 1155 determine whether it is abundant, deficient or perfect. 9 marks (iii) Show that every multiple of an abundant number is abundant. 4 marks (iv) Show that a power of a prime is deficient. 6 marks Solution to question 3: (i) Using the σ function from 5.2: σ(2 p 1 (2 p 1)) = σ(2 p 1 )σ((2 p 1)) = ( p 1 )(1 + 2 p 1) = 2 2 p 1 (2 p 1) so the proper divisors sum to 2 p 1 (2 p 1). (ii) 496 = 2 4 (2 5 1) by (i) is perfect. σ(945) = σ( ) = ( ) 6 8 = > so it is abundant. σ(1155) = σ( ) = < so it is deficient. (iii) Let n be abundant with proper positive divisors n 1,..., n r. So n n r > n. For a positive integer k we have kn kn r > kn but the kn i are some of the proper positive divisors of kn, so the sum of proper positive divisors of kn is strictly larger than kn. (iv) Let p be a prime then the proper positive divisors of p n are 1, p, p 2,..., p n 1 whose sum is (p n 1)/(p 1) < p n so it is deficient. 3

5 4(i) Find all the primitive Pythagorean triples which include the number marks (ii) You publish (n, e) = (1147, 11) in the RSA directory and receive the number 21. Decode it. 14 marks (iii) Determine the remainder when 99! is divided by marks Solution to question 4: (i) 2st = 2004 if and only if st = 1002 = in prime factors. Here are the triples: s t 2st s 2 t 2 s 2 + t (ii) 1147 = so we need d such that 11d 1 (modulo LCM(30, 36)) so 11d 1 (modulo 180). Using the Chinese Remainder Theorem or experimentation yields d = 131 because (11 131) 1 = 1440 = To decode we need (modulo 31) (21 30 ) 4 ( 10) = and (modulo 37) ( 3) ( 10) ( 11) 3 4. To solve x 4 (modulo 37) and x 12 (modulo 31) simultaneously use the CRT or search among x = 37k+4 12 for an integer divisible by 31. Either way one finds x 818 (modulo 1147) since 818 = 4+(37 22) = 12+(31 26). (iii) 103 is prime so 102! 1 (modulo 103) by Wilson s theorem. Hence 99! ( 1) ( 2) () 1 (modulo 103). Hence 99! is the solution to 6x 1 (modulo 103). By the Euclidean algorithm for LCM(6, 103) = 1 or experimentation one finds 6 86 = 516 = (5 103)

6 5(i). State Fermat s Little Theorem. 1 mark 5(ii) What day of the week will it be in days from today? 5 mark (iii) Find the remainder when is divided by marks (iv) Prove that a 561 a (modulo 561) for all integers a. 6 marks (v) The integer n > 1 is such that a n a (modulo n) for all integers a. Must n be prime? 2 marks (vi) The integer n > 1 is such that a n 1 1 (modulo n) for all positive integers a smaller than n. Prove that n must be prime. 5 marks Solution to question 5: (i) If p is a prime and a is an integer such that GCD(p, a) = 1 then a p 1 1 (modulo p). (ii) (modulo 7) so (modulo 7) so it is four days after today in the week. (iii) 561 = in prime factors we may use the CRT. Now (modulo 3), (3 10 ) 56 1 (modulo 11) and (3 16 ) 35 1 (modulo 17), Solve these three simultaneous congruences by the CRT algorithm or by experimentation to find that x 375 (modulo 561) is the solution. (iv) The smallest Carmichael number is n = 561 = Consider a 561 a (modulo 17), which is obvious if 17 a and otherwise we have a 561 = a (a 16 ) 35 a (modulo 17). If a 0 (modulo 3) then a 561 a (a 2 ) 280 a (modulo 3) and if a 0 (modulo 11) then a 561 a (a 10 ) 56 a (modulo 11), as required. (v) No - since 561 of (iv) is a counterexample. (vi) Suppose n is not prime. Then m n for some 1 < m < n. Therefore m n 1 1 (modulo n) and so 0 m n 1 1 (modulo m), which is a contradiction. 5

7 6(i) For each of the following numbers, find a prime number which divides it: , and marks (ii) Let p > 5 be a prime. Show that the congruence x 4 2x (modulo p) has no solutions if and only if p 2 or (modulo 15). 9 marks (iii) Solve the congruence of (ii) in the cases p = 29, 31, marks Solution to question 6: (i) 3 ( ) because (modulo 3) ( 1) ( ) because (modulo 17) ( ( 1) is a prime so, by Euler s criterion, 2 83 ( ) (modulo 167), since (modulo 8) or because (modulo 167). (ii) x 4 2x 2 15 = (x 2 5)(x 2 +3) so x is a solution of the congruence if and only if x 2 3 (modulo p) or x 2 5 (modulo p). Hence there is no solution if and only if ( ) ( ) ( ) ( ) 3 p = 1 = 5 p. This is equivalent to p 3 = 1 = p 5, by Quadratic Reciprocity and ( ) 1 p = ( 1) (p 1)/2. This happens if and only if p 2 (modulo 3) and p 2 or 3 (modulo 5), which happens if and only if p 2 or (modulo 15). (iii) p = 29 : By (ii) there is no solution to x (modulo 29) but x = ±11 (modulo 29) is a solution to x p = 31 : By (ii) there are 4 solutions and these are x ±11, ±6 (modulo 31). p = 37 : By (ii) x 2 5 has no solution but x = ±16 (modulo 37) satisfies x

8 7.(i) Define a perfect number. 1 mark Calculate and Let n be a perfect number. State the value of d n 1 where the sum is over all the positive d divisors of n. 3 marks Deduce that no proper divisor of a perfect number is perfect and no proper multiple of a perfect number is perfect. 5 marks (ii) Two positive integers are said to be amicable numbers if each is the sum of the proper divisors of the other. The prime numbers p, q, r are such that p = (3 2 n ) 1, q = (3 2 n 1 ) 1 and r = (9 2 2n 1 ) 1 for some integer n > 1. Use the result that σ(ab) = σ(a)σ(b) if GCD(a, b) = 1 to prove that σ(2 n pq) = σ(2 n r) = 2 n pq + 2 n r. 10 marks Deduce that 2 n pq and 2 n r are amicable. 4 marks Illustrate with an example 2 marks Solution to question 7: (i) A perfect number is a positive integer equal to the sum of its proper 1 positive divisors = 2 and = 2. If n is perfect d n 1 = 2. d Let n be perfect and m n with 1 < m < n and let r > 1 then d m 1 < d d n 1 = 2 and d d rn 1 > d d n 1 = 2 so m and rn are not perfect. d (ii) σ(2 n pq) = σ(2 n )σ(p)σ(q) = ( n ) 3 2 n 3 2 n 1 = (2 n+1 1) 9 2 2n 1 and σ(2 n r) = (2 n+1 1) 9 2 2n 1, as required. Also 2 n pq + 2 n r = 2 n ((3 2 n ) 1)((3 2 n 1 ) 1) + ((9 2 2n 1 ) 1)) = 2 n ( n n 1 (2 + 1)) = (2 n+1 1) 9 2 2n 1. The sum of the positive proper divisors of 2 n pq is σ(2 n pq) 2 n pq = 2 n r and the sum of the positive proper divisors of 2 n r is σ(2 n r) 2 n r = 2 n pq so 2 n pq and 2 n r are amicable. Putting n = 2, p = 11, q = 5, r = 71 we see that 2 n pq = 220 and 2 n r = 284 are amicable. 7