Solutions to Problems; Chapter (^4 321)
The result can be generalized to polynomials of degree n. See Problem
4426 in Amer. Math. Monthly 58 (1951), 113; 59 (1952), 419.
9.7. Let n = 1 and p(x) = az + b. For m = 0, the condition is that C~CJZ +
(cc + cib) = 0 identically; f or m = 1, the condition is that (cc + ci)ae +
(cc+ci)b+cia = 0 identically. In either case, it can be seen that cc = cl = 0.
Suppose that the result holds for polynomials of degrees up to n - 1
inclusive. Let degp(z!) = n and 0 5 m _< n - 1. Then, differentiating the
condition leads to a similar condition for the derivative p’(z) and the result
follows in a straightforward way from the induction hypothesis.
The only case left to consider is that
cop(z) + c1p(2 + 1) + * + cnP(z + n) = O
identically. Since the leading coefficient must vanish, cc + cl +... + c, = 0,
whence
0 = c&(x + 1) -p(z)] + c&(c + 2) - P(Z)] +... + Glb(2 + n> - P(Z)1
= (Cl + c2 +... + cn)q(z> + (cz +... + G$?(~ + 1)
- (c3+. f + c)q(z + 2) +... + c,q(z + n - 1)
where q(z) = p(z + I) - p(z), a polynomial of degree n - 1. The desired
result now follows from the induction hypothesis.
9.8. Let
f(t) = &(t - rip,
i=l
where ml + rns +... + ?nk = n. Then
f'(t) = Cfi(t - Ti)"i-lg(t)
i=l
for some polynomial g(t) which does not vanish at any ri. The degree of
(n - 1) - &(mi - 1) = L - 1.
i=l
But since a power of f’(t) divides a power of f(t), each zero of g(t) is a zero
of f(t). But, since g(t) and f(t) do not share any zero, g(t) is constant, so
that L = 1, ml = n. It is easily checked that any polynomial of the form
c(t - T)” satisfies the conditions of the problem.
9.9. Suppose that n = degp(t) 2 degq(t) and that, if possible, (p - q)(t)
is not identically zero. Let the distinct zeros of p(t) and q(t) be ~1,... , U,