Solutions to Problems; Chapter (^8 381)
Since this guarantees that either 1 is a zero or f(1) and f(-1) have oppo-
site signs, he can win regardless of the last move of the first player. If s is
odd, he needs to take a little more care. In this case, he notes that
2’f(l) + f(-2) = 2’g(l) + g(-2) + (2’ - 2”)b.
Again, he can choose b to ensure that this quantity vanishes regardless of
the first player’s last move and thus, as before, ensure a win.
- For the case m = n = 4, the strategy is to raise AX - BY to a
sufficiently high power that A4 or B4 is a factor of each term. Thus,
1 = (AX - BY)7 = A4(A3X7 - 7A2BX6Y + 21AB2X5Y2 - 35B3X4Y3)- B4(B3Y7 - 7AB2XY6 + 21A2BX2Y5 - 35A3X3Y4).
The choice of u and v is clear. For general m, n, look at 1 = (AX -
BY) m+n-1.- Let q(x) = p(x) + p’(x) + ee. + p(“)(x). Then q(x) is a polynomial
of even degree with positive leading coefficient. Suppose q(x) assumes its
minimum value when x = u. Then 0 = q’(u) = q(u) - p(u). But then
q(x) 2 q(u) = p(u) 2 0 for all z (cf. Problem 5.4.22). - Suppose a 5 b 5 c. Let f(t) = (t - x)(t - y)(t - Z) and g(t) =
(t - u)(t - b)(t - c). S ince f(c) 1 0 = g(c) and f(a) < 0 = g(u), there exists
some point u in the interval [a,~] for which f(u) = g(u). This implies that
(xy + xz + yz - ab - ac - bc)u = 0. Since u 1 a > 0, it follows that
xy + xt + yz = ab + UC + bc and f(t) = g(t) identically. The result follows.
For other solutions, see Amer. Math. Monthly 72 (1965)’ 185-186. - Let n be a given integer. We can write the given polynomial in the
form p(x) = k fl(x - ri). Let
q(x) = xn JJ(xflwl + riXne2 +... + ry-“x + rr-‘).Then p(x)q(x) = Lx” n(xn - ry). We obtain the result by taking n =
1 000 000. [Solution by M.S. Klamkin.]
- Suppose that the degree of p(x) is m and that of q(x) is n where
m 1 n. If the distinct zeros of p(x) and q(x) are ui, 212,... , u, and the
zeros of p(x) - 1 and q(z) - 1 are vi, ~2,... , v6, then the ui and the vj are
distinct zeros of the difference (p - q)(z). If we can show that r + s > m >
deg(p - q)(z), the result will follow.
Now, the common derivative of p(x) and p(x) - 1 has at least (m - r) +
(m - s) zeros counting multiplicity, so that 2m- r - s 5 degp’(z) = m - 1,
whence r + s > m + 1 > m.