Concrete mathematics : a foundation for computer science

530 ANSWERS TO EXERCISIES
6.40 If 6n - 1 is prime, the numerator of
4n ’ (-‘)k- ’
tjy-= H4n 1 -Hzn I
k=l
is divisible by 6n - 1, because the sum is
Similarly if 6n + 1 is prime, the numerator of x”,E, (-1 )km ‘/k = Han ~ Hln
is a multiple of 6n + 1. For 1987 we sum up to k = 1324.
6.41 ‘&+I = tk (Lin+‘k+kl’LJ) = tk (L’“~k~‘2J), hence we have Sn+l + S, =
xk(
6.42 F,,.
11*-i kL’/2+11) = Sn+2. The answer is F,+z.
6.43 Set z = $ in Ena0 F,z" = z/(1 - z - z2) to get g. The sum is a
repeating decimal with period length 44:
0.1123595505617977ti28089887640449438202247191011235955+
6.44 Replace (m, k) by (--m, -k) or (k, -m) or (--k, m), if necessary, so
that m 3 k 3 0. The result is clear if m = k. If m > k, we can replace (m, k)
by (m - k, m) and use induction.
6.45 X, = A(n)oc+B(n)fi-tC(n)-y+D(n)&, where B(n) = F,, A(n) = F, 1,
A(n) + B(n) - D(n) = 1, and B(n) - C(n) + 3D(n) = n.
6.46 $/2 and @ -l/2. Let LL = cos 72” and v = cos 36”; then u = 2v2 - 1 and
v = 1-2sin' 18” = 1-2~‘. Ijence u+v = Z(u+v)(v-u), and 4v2-2v-1 = 0.
We can pursue this investigation to find the five complex fifth roots of unity:
1,
Q-1 f i&Gfl -Q f i&q
2 ’ 2
6.47 2Q5 F, = (1 + &)” - (1 - &)n, and the even powers of fi cancel "Let p be any old
out. Now let p be an odd prime. Then (2kF,) = 0 except when k = (p - 1)/2, prime.”
(See j140/, p. 419.)
;;d (f,,!,) = 0 except when k = 0 or k = (p - 1)/2; hence F, E 5(p ‘)/’ and
p+l E 1 +5(Pm')/' (mod p). It can be shown that 5(PP’)/2 E 1 when p has
the form ‘Ok & 1, and 5(P ’ :/2 E -1 when p has the form ‘Ok f 3.
6.48 This must be true because (6.138) is a polynomial identity and we can
set a, = 0.