Unknown

Solutions to Problems; Chapter 7 367

which contradicts the induction hypothesis.

5.7. Since only differences of the ni are involved, it suffices to obtain the
result when all the ni are nonnegative. Let

G(x) = G(xo, xl,... , Xk) = I-I (7; 1;;’

both products taken over 0 5 i < j 5 k. We have to show that G takes
integer values whenever x = n is a vector with integer entries. This will be
proved by induction on n = max(ns, nl,... , nk).
If n 5 k - 1, since all the ni are nonnegative and there are k + 1 of them,
two of the ni must be equal and G(n) = 0. If n = k, then either two ni are
equal or else {no, nl,... ,nk} = {0,1,2,... , k}. In either case, G(n) is an
integer.
Suppose it has been shown that G(n) is an integer when n 5 r for some
T > k. Let n = r + 1. Without loss of generality, we may suppose that
no = P + 1. If, for some i > 0, ni = P + 1, then G(n) = 0. Suppose that
1zi 5 r (1 5 i _< r). Let

f(t) = G(t, m, n2,... , ~1.

By the induction hypothesis, f(t) is an integer for t = 0, 1,2,... , k,... ,r.
Since deg f(t) = k, it follows from Exercise 1.19, that f(r+ 1) is an integer.
Hence G(n) is an integer if n = r + 1.

5.8. Let n be a positive integer to be determined later, and let p(x) =
0.5[1+ (2x - l)n]. s ince all the coefficients of 1 + (2x - 1)” are even, p(x)
has integer coefficients. Also

Ip(x) - 0.51 = 0.512~ - I]“.

If 0.19 5 x 5 0.81, then lp(z)-0.51 5 0.5(0.62)“. Now choose n sufficiently
large that (0.62)n < 2/1981.

5.9. First solution. In a diagram, let the graph of the function dm
be given. Let us choose p and q so that the maximum value of the left
side over all x in [0, l] is made as small as possible. From the diagram, we
see that the line with equation y = px + q should go through the points
(0, 1 + u) and (1,~) f or some positive value of u. Hence, slope p of the line
must be -1 and it must also go through (l/J”, l/a-u). This is possible
only if q = (l/2)(1 + 4) and u = (l/2)(4 - 1). From this, we see that
the inequality of the problem is always satisfied if and only if p = 1 and
q = (l/2)(1 + 4).
Second solution. Let f(z) = dm - pz - q (0 5 x 5 1). We first
establish that p has to be negative. For,

f(0) = 1 - q 5 (l/2)@- 1) * -q 5 (l/2)@- 1) - I