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20. Find the value of (x - a) + (x – b)" + (x – c) – 3 (x – a) (x - b) (x – c), when
a + b + c = 3r.
Answers
Answers and Solutions
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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