4. Let P = {x Xe N and x is a factor of 20} and Q = {x | Xe N and x is a factor of 32).
Then:
(i) find n(P) + 2 x n(Q)
(ii) find 3 x n(P) + 4 x n(Q)
Answers
Answer:
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 = 2n2, and the result holds.
3.2(a). 2mn - (m2 - n2 ) = 1 can be rewritten (m + n)2 - 2m2 = 1, while
(m2 - 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 - [a2 + (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 - (m2 - nZ) = 7 can be rewritten (n + m)2 - 2m2 = 7, and
(m2 - 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) = (m2 - 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)].