3.2.2 ANSWERS TO (How to cite a web site) EXERCISES 527 X, 3 cn
3.2.2 ANSWERS TO EXERCISES 527 X, 3 cn + pY, (modulo p ), we must have Yn+l G n2c2r + ncs + Y, (modulo p); hence Y, G (:)2 c2r + (;)(c ~ + cs) (modulo p). Thus YP modp = 0, and the desired relation has been proved. Now we can prove that the sequence (Xtl) of integers defined in the hint satisfies the relation X n+pf = X, + tpf (modulo pf+ ), n 2 0, for some t with t mod p # 0, and for all f 2 1. This suffices to prove that the sequence (X, modp ) has period length p , for the length of the period is a divisor of pe but not a divisor of pe- . The above relation has already been established for f = 1, and for f > 1 it can be proved by induction in the following manner: Let X nfpf -xn + tpf + Znpffl (modulo pffz); then the quadratic law for generating the sequence, with d = pr, a = 1+ ps, yields z n+l -2rtnc + st + 2, (modulo p). It follows that Zn+P E Z, (modulo p); hence X n+kpf = X, + k(tpf + Z,pf+ ) (modulo pff2) for k = 1,2,3,. . ; setting k = p completes the proof. Notes: If f(z) is a polynomial of degree higher than 2 and Xn+l = f(Xn), the analysis is more complicated, although we can use the fact that f (m + pk) = f(m) + pk f (M) + p2 f (m)/2! + . . . to prove that many polynomial recurrences give the maximum period. For example, Coveyou has proved that the period is m = 2 if f(0) is odd, f’(j) 3 1, f”(j) -0, and f(j+ 1) = f(j)+ 1 (modulo 4) for j = 0,1,2,3. [Studies in Applied Math. 3 (1967), 70-111.1 9. Let X, = 4Y,+2; then the sequence Y, satisfies the quadratic recurrence Y,+l = (4Yi + 5Y, + 1) mod ae- . 10. Case 1: XO = 0, X1 = 1; hence X, = F,. We seek the smallest n for which F,, s 0 and F,+I G 1 (modulo ae). Since Fzn = Fn(Fn-l + Fn+l), Fzn+l = Ft + Fi+l, we find by induction on e that, for e > 1, F3.2e–1 = 0 and F3.2c-~+1 SE 2 + 1 (modulo 2 + ). This implies that the period is a divisor of 3 . Ze- but not a divisor of 3 + 2e-2, so it is either 3 . Ze- or 2e-1. But Fze-1 is always odd (since only Fan is even). Case 2: X0 = a, X1 = b. Then X,, = aF,-1 + bF,; we need to find the smallest positive n with a(F,+l -Fn) + bF, ZE a and aF, + bF,+l E b. This implies that (b2 -ab -a )F,, z 0, (b2 -ab - a2)(F n+l -1) E 0; and b2 -ab - a2 is odd (i.e., prime to m) so the condition is equivalent to F, z 0, F,+I G 1. Methods to determine the period of F, for any modulus appear in an article by D. D. Wall, AMM 67 (1960), 525-532. Further facts about the Fibonacci sequence mod 2 have been derived by B. Jansson [Random Number Generators (Stockholm: Almqvist & Wiksell, 1966), Section 3Cl]. 11. (a) We have zx = 1 + f(z)u(z) + p v(z) for some u(z), V(Z), where V(Z) $ 0 (modulo f(z) and p). By the binomial theorem zxp = 1 + pe+ w(z) + p + V(z)2(p -1)/2 plus further terms congruent to zero (modulo f(z) and pe+ ). Since pe > 2, we have zxP E I+$+ V(Z) (modulo f(z) and p + ). If peflv(z) = 0 (modulo f(z) and pef2),
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