Answers and Solutions Chapter 1 1.1(a). The point E can be obtained by folding the square along the bisector of angle BAC, so that AB is folded onto the diagonal with B landing on E. If F is the point where the bisector intersects B C, then B F folds to E F. Thus E F -1 AC and BF = EF. Since L.EFC = 45° = L.ECF, FE = EC. l.1(b). WC! = IBC! - IBFI = IAEI - ICEI = 21AEI - lAC! and IEC! = lAC! - IAEI = lAC! - IBC!. l.3(a). n Pn qn rn 1 1 1 1 2 3 2 3/2 = 1.5 3 7 5 7/5 = 1.4 4 17 12 17/12 = 1.416666 5 41 29 41/29 = 1.413793 1.3(c). Clearly, rn > 1 for each n, so that from (b), rn+1 - rn and rn - rn-I have opposite signs and 1 Irn+l - rnl < 41rn - rn-II· Since r2 > 1 = rl, it follows that rl < r3 < r2. Suppose as an induction hypothesis that rl < r3 < ... < r2m-1 < r2m < ... < r2. Then 0 < r2m - r2m+1 < r2m - r2m-1 and 0 < r2m+2 - r2m+1 < r2m - r2m+J. so that r2m-1 < r2m+1 < r2m+2 < r2m. Let k, 1 be any positive integers. Then, for m > k, I, r2k+1 < r2m-1 < r2m < r21. l.3(d). Leta be the least upper bound (i.e., the smallest number at least as great as all) of the numbers in the set {rl, r3, r5, ... , r2k+l, ... }. Then r2m-1 ::::: a ::::: r2m 139 140 Answers and Solutions .....-_______---,D \ \ \ \ \ \ \ \ \ \ BL..---...L..-----~ F FIGURE 1.3. for each positive integer m, so that la - r2m-ll and la - r2m I are both less than 1 1 Ir2m - r2m-ll .:::: 42(m-l) Ir2 - rll = 24m - 3 · It follows that limn--->oo rn = limn--->oo rn-l = a. Taking limits in the recursion in (d) leads to 1 a=I+-- l+a which reduces to a 2 = 2. Since rn > 0 for each n, a > O. 1.4(c). 347/19 = 18 + 1/3 + 1/1 + 1/4. 1.4(d). Since v'2 = 1 + (v'2 - 1) = 1 + 1/1 + v'2, it follows that the process does not terminate, since we can always replace the final 1 + v'2 by 2 + 1/1 + v'2 to get the continued fraction in Exercise 1.3. 1.7(b). This can be established from Exercise 1.5(a) by induction. 1.7(c). For each positive integer k, let P2k = Xk and q2k = 2Yk. Then, from Exercise 1.5(b), xf - 8Yf = P~k - 2q~k = 1. 1.7(d). If X2 - 8y2 = -1, then x would have to be odd. But then x2 == 1 (mod 8), yielding a contradiction. 2.4(a). Since qn + qn-l = (qn - qn-l) + 2qn-l = Pn-l + (Pn - Pn-l) = Pn, Pn(qn - qn-l) = PnPn-l = Pn-l (qn + qn-d. 2.4(b). From Exercise 2.3(c), Pn+lqn+l = 4pnqn + 2(Pnqn-l + Pn-lqn) + Pn-lqn-l = 4pnqn + 2(Pnqn - Pn-lqn-d + Pn-lqn-l = 6pnqn - Pn-lqn-l· 3.1(b). Yes. It suffices to consider the case for which the greatest common divisor of a, b, cis 1. Since each odd square leaves a remainer 1 and no square leaves a Chapter 1 141 remainder 2 upon division by 4, both of a and b cannot be odd. Suppose b is even and a is odd. Then b2 = (c - a)(c + a) expresses b as the product oftwo even divisors whose greatest common divisor is 2. Thus, there are integers m and n for which c + a = 2m2 and c - a = 2n 2, and the result holds. 3.2(a). 2mn - (m 2 - n2 ) = 1 can be rewritten (m + n)2 - 2m2 = 1, while (m 2 - n2) - 2mn = 1 can be rewritten (m - n)2 - 2n2 = 1. We obtain Pell's equation x2 - 2y2 = 1, where (x, y) = (m + n, m), (m, n) = (y, x - y), and (x, y) = (m - n, n), (m, n) = (x + y, y). 3.2(b). (m, n) = (2, 1) gives the triple (3, 4, 5), while (m, n) = (5,2) gives the triple (21, 20, 29). 3.2(c). For example, (x, y) = (17,12) leads ultimately to the triples (119, 120, 169) and (697, 696, 985). 3.3. (a + c)2 + (a + c + 1)2 - (2a + c + 1)2 = 1 - [a 2 + (a + 1)2 - c2] . 3.4(a). q~+l - q~ - 2qn+lqn = (qn+l - 2qn)qn+l - q~ = qn-lqn+l - q~ = qn-l (2qn + qn-l) - qn (2qn-l + qn-2) 2 = qn-l - qnqn-2 = ... = (-W(q? - qoqZ) = (-It. 3.5. 2mn - (m 2 - nZ) = 7 can be rewritten (n + m)2 - 2m 2 = 7, and (m 2 - n2) - 2mn = 7 as (m - n)2 - 2n2 = 7. The equation x2 - 2y2 = 7 is satisfied by (lxi, Iyl)] = (3,1), (13,9), (75,53), and these yield the triples (-3, 4, 5), (8, 15, 17), (65, 72, 97), (396, 403, 565), (2325, 2332, 3293), (13568, 13575, 19193). 3.6. Let a2 + b2 = c2, p2 + q2 = r2 with p = a + 3, q = b + 3, and r = c + 4. Then we must have 3a + 3b + 1 = 4c. Taking (a, b, c) = (m 2 - n2, 2mn, m2 + n2), we are led to (m - 3n)2 - 2n2 = 1. Suppose that x = m - 3n and y = n. Then x 2 - 2l = 1. Some solutions are (x, y) = (3,2), (-3,2), (17,12), (-17,12), which give rise to (m, n) = (9,2), (3,2), (53, 12), (19, 12). Thus we have some examples: [(a, b, c), (p, q, r)] = [(77,36,85), (80,39,89)], [(5,12,13), (8, 15, 17)], [(2665, 1272,2953), (2668, 1275,2957)], [(217,456,505), (220,459,509)]. 142 Answers and Solutions Comments on the Explorations Exploration 1.2. If X2 - 2y2 = ±2, then x would have to be even, say x = 2z, so that y2 - 2z2 = =fl. The equation x 2 - 2y2 = 2 is satis­ fied by (x, y) = (2,1), (10,7), (58,41), and x 2 - 2y2 = -2 by (x, y) = (4,3), (24, 17), (140,99). As for x 2 - 2y2 = 3, if y is even, then x 2 == 3 (mod 8), while if y is odd, then x 2 == 5 (mod 8); neither congruence for x 2 is realizable. A similar argument shows the impossibility of x 2 - 2y2 == -3. Exploration 1.3. We have to solve the equation x 2 - 3y2 = 1, where x = 2u is even and y = 2 v + 1 is odd. The following table gives the first few triangles of the required type: (x, y) (v, v + 1, u) (26, 15) (7,8, 13) (362,209) (104, 105, 181) (5042,2911) (1455, 1456,2521) (70226, 40545) (20272,20273,35113) (978122,564719) (282359,282360,489061) Observe that the equation satisfied by u and v implies that (v + 1)3 - v 3 = u2• It is known from this that u must be the sum of two consecutive squares. (See Power Play by E.J. Barbeau (MAA, 1997) pages 7,8, 11 for a proof of this.) 13 = 22 + 32 , 181 = 92 + 102, 2521 = 352 + 362, 35113 = 1322 + 1332, 489061 = 4942 + 4952. Adding together the consecutive roots of the squares in each right side gives the sequence {5, 19,71,265,981, ... }, which satisfies the recursion tn+l = 4tn - tn-I. The terms of the sequence are sums of the consecutive terms of a second sequence {l, 4, 15, 56,209, ...} that satisfies the same recursion. The values of y when x 2 - 3 y2 = 1 are not only found among its terms, but the numbers x and y are intricately related in other ways. For example, (5042, 2911) = (71 2 + 1, 562 -15). Is there something here that can be generalized? Exploration 1.4. This exploration can be approached in various ways. For x 2 + mx ± n to be factorizable, the discriminants m 2 =f 4n must be integer squares, say m2 - 4n = u2 and m2 + 4n = v2. In particular, we need u2 + v2 = 2m2. We can achieve this if m2 = p2 + q2. Then 2m2 = (p + q)2 + (p - q)2, so we can try u = p - q and v = p + q. This leads to n = pqj2. Thus we can find polynomials involving any number m whose square is the sum of two squares. Since m is thus the largest of a Pythagorean triple, we can let m = y (a2 + f32), Chapter 1 143 p = y (a2 - f32), and q = 2yaf3. Then n = pq 12 = y 2af3(a2 - f32). This leads to x 2 + mx + n = (x + ya(a - f3))(x + yf3(a + f3)), x 2 + mx - n = (x + ya(a + f3))(x - yf3(a - f3)). Alternatively, we need to find numbers a, b, e, d for which (x + a)(x + b) = (x - e)(y + d). This yields m = a + b = d - e and n = ab = cd. Substituting d = able into the first equation gives e2 + (a + b)e - ab = 0 with discriminant (a + 3b)2 - 8b2. If this is the square of w, then 8b2 = (a + 3b)2 - w2 = (a + 3b + w)(a + 3b - w). To construct possibilities, let b be chosen arbitrarily and write 8b2 = (2r) (2s). Then solving a + 3b + w = 2r, a + 3b - w = 2s gives a + 3b = r + sand w = r - s, leading to the possibilities (a, b, e, d) = (r + s - 3b, b, b - s, r - b), (m, n) = (r + s - 2b, b(r + s) - 3b2). Thus x 2 + mx + n = (x + b)(x + r + s - 3b) and x 2 + mx - n = (x + s - b)(x + r - b).