Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 505
4.45 x2 E x (mod 10n) ( x(x - 1) E 0 (mod 2”) and x(x - 1) E 0
(mod 5n) M x mod 2” = [Oor 11 and x mod 5” = [Oor 11. (The last step
is justified because x(x - 1) mod 5 = 0 implies that either x or x - 1 is a
multiple of 5, in which case the other factor is relatively prime to 5n and can
be divided from the congruence.)
So there are ,at most four solutions, of which two (x = 0 and x = 1)
don’t qualify for the title “n-digit number” unless n = 1. The other two
solutions have the forms x and 1 On + 1 -- x, and at least one of these numbers
is > 1 On-‘. When n = 4 the other solution, 10001 - 9376 = 625, is not a
four-digit number. 1Ne expect to get two n-digit solutions for about 90% of
all n, but this conjecture has not been proved.
(Such self-reproducing numbers have been called “automorphic.“)
4.46 (a) If j’j - k’k = gcd(j,k), we have nk’knscdii,k) = ni’i = 1 and
nk’k - 1. (b) L e tn
= pq, where p is the smallest prime divisor of n. If
2” E 1 (mod n) then 2” G 1 (mod p). Also 2P-l = 1 (mod p); hence
2scdipm ‘,nl = 1 (mod p). But gcd(p - 1 In) = 1 by the definition of p.
4.47 If n+’ = 1 (mod m) we must have n I m. If nk = nj for some
1 < j < k < m, then nkPj = 1 because we can divide by nj. Therefore if the
numbers n’ mod m, . , n”-’ mod m are not distinct, there is a k < m - 1
with nk = 1. The least such k divides m- 1, by exercise 46(a). But then kq =
(m - 1 )/p for some prime p and some positive integer q; this is impossible,
since nkq $ 1. Therefore the numbers n’ mod m, . . , nmP’ mod m are
distinct and relatively prime to m. Therefore the numbers 1, . , m - 1 are
relatively prime to n-L, and m must be prime.
4.48 By pairing numbers up with their inverses, we can reduce the product
(mod m) to n l~n 2); otherwise it’s +l.
4.49 (a) Either m < n (@(N - 1) cases) or m = n (one case) or m > n
(O(N - 1) again). Hence R(N) = 2@(N - 1) + 1. (b) From (4.62) we get
2@(N-l)+l = l+x~(d)LN/d]LN/d-11;
d>l
hence the stated result holds if and only if
x p(d) LN/dJ = 1 ,
d2’
for N :z 1
And this is a special case of (4.61) if we set f(x) = (x 3 1)