Large MA132 Foundations Hints and Solutions to ... - Of the Clux

1 Large MA132 FoundationsHints and Solutions to Exercises II1. Find the hcf and lcm of (i) 10 6 and 10 6 + 10 3 (ii) 90090 and 2200. (iii) 10 6 + 144and 10 3 − 12 (iv) 90090, 2200 and 1525.Solution: (i) hcf = 10 3 , lcm = 10 3 × (10 6 + 10 3 ). (ii) hcf = 110, lcm = 1801800 (iii)hcf = 4, lcm = (106 +144)(10 3 −12)(iv) hcf = 5, lcm = 109909800.43. There is a way of finding the hcf of three numbers by first finding the hcf’s of pairs ofthem. What is it? How would you find the hcf of four very large (i.e. too large to factorise)numbers?Solution: hcf(m, n, p) = hcf(hcf(m, n), p); that is, find the hcf of the first two, then find thehcf of that and the third number. Similarly, hcf(m, n, p, q) = hcf(hcf(m, n), hcf(p, q)). Usethe Euclidean algorithm for each calculation of hcf.4. Draw diagrams representing all the divisibility relations between the numbers 1, 2, 4, 3, 6, 12,and all the inclusion relations between the subgroups Z, 2Z, 4Z, 3Z, 6Z, 12Z.Solution:12 6 4 3 2 112Z 6Z 3Z 2Z Z4Z 5. In the lectures we proved that if m and n are any two integers then there are integers aand b such that am + bn = hcf(m, n). Find a and b when m and n are1. 365 and 748 (from Example 3.9(i) in the Lecture Notes)2. 365 and 750 (from Example 3.9(ii) in the Lecture Notes)3. 10 6 + 144 and 10 3 (from Exercise 3 above)4. 90, 090 and 2, 200 (from Exercise 3 above).Hint: keep a careful record of your use of the Euclidean algorithm when you find the hcf’s.Solution for 365 and 748: Performing successive divisions, we find748 = 2 × 365 + 18 = 142 × (748 − 2 × 365) − 7 × 365 =365 = 20 × 18 + 5 = 2 × 18 − 7 × (365 − 20 × 18)18 = 3 × 5 + 3 = 2 × (18 − 3 × 5) − 1 × 55 = 1 × 3 + 2 = 1 × 3 − (5 − 1 × 3)3 = 1 × 2 + 1 1 = 3 − 2

Hence hcf(365, 748) = 1, and using the last calculation from the bottom up,1 = 1 × 3 − 1 × 2= 1 × 3 − (5 − 1 × 3)= 2 × 3 − 1 × 5= 2 × (18 − 3 × 5) − 1 × 5= 2 × 18 − 7 × 5= 2 × 18 − 7 × (365 − 20 × 18)= 142 × 18 − 7 × 365= 142 × (748 − 2 × 365) − 7 × 365= 142 × 748 − 291 × 365where each appearance of a subtraction in parentheses indicates that we have used one ofthe lines in the previous calculation to make a substitution.6.Solve the equation x 2 + y 2 = 1 in integers.Solution: (x, y) = (±1, 0), (0, ±1).7. If x and y are integers and x 2 + 6 = y 3 ,(i) Show that x is odd.Solution: If x is even, write x = 2w and get 4w 2 + 6 = y 3 . Either y is even or y is odd. Inthe first case y 3 is divisible by 2 3 and hence by 4. The equation 4w 2 + 6 = y 3 then implies6 is divisible by 4, an absurdity. If y is odd then so is y 3 . But the equation 4w 2 + 6 = y 3implies y 3 is even.8.(i) Suppose x, y, z are non-zero integers satisfyingShow that x, y and z are all even.x 3 + 2y 3 + 4z 3 = 0.Solution: x can’t be odd, for if it were then x 3 would be odd, but the equation x 3 + 2y 3 +4z 3 = 0 shows this is impossible. Write x = 2x 1 and substitute, to get 8x 3 1 + 2y 3 + 4z 3 = 0.Divide throughout by 2, and now apply a similar argument to y . . . .(ii)(Harder) Show that there are no such integers x, y, z.Solution: Let 2 k be the highest power of 2 dividing each of x, y, z. Then each of x/2 k , y/2 k , z/2 kis an integer, and one at least is odd. Divide the equation x 3 + 2y 3 + 4z 3 = 0 by 2 3k , getting(x/2 k ) 3 + (y/2 k ) 3 + (z/2 k ) 3 = 0. Now apply (i) to get a contradiction.9. Show that if x and y are non-zero integers and x 2 = y 3 then x = a 3 and y = b 2 for someintegers a and b.Solution: Use prime factorisations of x and y: x = p a 11 × · · · × p ann , y = p b 11 × · · · × p bnn . Theequation x 2 = y 3 implies that 2a j = 3b j for all j. So b j is even and . . .

310. Show that if hcf(x, y) = 1 and xy = a 2 then x = a 2 1 and y = a 2 2 for some integers a 1 anda 2 .Solution: Use prime factorisation of x and y again.11. List the squares mod 4. Show that x 2 + 2 ≠ 5y 2 for any integers x, y.Solution: 0 2 = 0 mod 4, 1 2 = 1 mod 4, 2 2 = 0 mod 4, 3 3 = 1 mod 4, 4 2 = 0 mod 4, 5 2 =1 mod 4, . . . Do you see the pattern? Prove that the pattern you see is really there. Nowyou’ve done the first part of the question. If x 2 + 2 = 5y 2 then x 2 + 2 = y 2 mod 4. Lookback at the list of squares mod 4. Moral (if you didn’t succeed with this question until yourlooked at this solution): begin by doing the simplest thing possible in the circumstances. Inthis case, finding the squares of the first few natural numbers mod 4 reveals a pattern thatis not difficult to prove.12. Adapt the argument, used in the Lecture Notes to prove that √ 2 /∈ Q, to prove that ifn ∈ N is not a perfect square (i.e. not the square of a natural number), then √ n /∈ Q.Solution: That n is not a perfect square means that in its prime factorisation, some primep appears raised to an odd power. Suppose this power is 2k + 1, so thatn = p 2k+1 n 1 with p not dividing n 1 . (1)If √ n = r s with r, s coprime integers, then ns2 = r 2 . Hence p 2k+1 divides r 2 . As the powerof p in the prime factorisation of r 2 is even, it must be at least 2k + 2, so p k+1 |r, and we canwrite r = p k+1 r 1 . Combining this with (1) we find thatsop 2k+1 n 1 s 2 = p 2k+2 r 2 1;n 1 s 2 = pr 2 1.As p does not divide n 1 , it must divide s 2 , and therefore it divides s. But now p|r and p|s,contradicting the fact that r and s are coprime.