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376 Answers to Exercises and Solutions to Problems

Second solution. Since p(x) and q(x) are continuous, then p(x) > q(x)

for all x or p(x) < q(x) for all x. Suppose the former. Then p(p(x)) >

q(p(x)) = p(q(x)) > q(q(x)) for all x and the result follows.

20. Suppose, first, that AB and AC lie along the lines x = b and y =
    c. Then the polynomial P(b, y) has three distinct zeros and so vanishes
    identically. Similarly, P(x, c) must be the zero polynomial. Hence P(x, y) =
    k(x - b)(y - c). Since A’ lies on neither AB nor AC, P(z, y) vanishes at
    the coordinates of a’ only if k = 0.
    Otherwise, we may suppose that the lines AB, AC and BC lie along the
    respective lines y = mix+ki (i = 1,2,3). Since for each i, P(x, mix+ki) = 0
    has three distinct roots, the left side must be the zero polynomial. But then,
    by the Factor Theorem, P(x, y) is divisible by the three distinct factors
    y - mix - ki. Since deg P(x, y) 5 2, this is possible only if P(x, y) = 0.


21. By the Factor Theorem, (ca - b)(b - c)(c - a) divides both the nu-
    merator and the denominator. Arguing from degree and coefficients, the
    denominator must be equal to this and the numerator must have an addi-
    tional symmetric linear factor, so that the result follows.


22. Let u = (3/5)‘j7, v=~+u-~.Thenu~+u-~=v~-3v,

u5 + u -5 = v5 - 5(v3 - 3v) - 1ov = v5 - 5v3 + 5v

and

u7 + u-7 = v7 - 7(v5 - 5v3 + 5v) - 21(v3 - 3v) - 35v

=v 7 - 7v5 + 14v3 - 7V’

whence v is a zero of the polynomial

t7 - 7t5 + 14t3 - 7t - (3/5 + 5/3).

Multiplying this by 15 yields the required polynomial.

23. (a) P(X) = l-n(l-zai/ni), f rom which the result follows immediately.
    (b) L e t r - - g c 4 m, n) and d be the number of prime factors counting
    repetitions of r. Then m = ru, n = rv, k = ruv, c = a + b - d and

p(x)=< [l- (1-T) (l-q-f)].

If, say, m = 2”, then r = 2d and p(2) = 1. On the other hand, if p(2) = 1,
then 2d/r must equal 1 and either 2a-d = u or 2b-d = v.

Remark. A generalized version of this problem in which there are s
ni not pairwise relatively prime is found in the Canadian Mathematical
Bulletin 24 (1981), 507, P292.