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Solutions to Problems; Chapter 2 269

5.6.

(r + l)Sr + ( ’ i ’ ) Srel +.. - + (r + 1)Si = k k ( ’ f ’ ) k’+l-j

k=l j=l

= k[(l + k)‘+l - k’+’ - l] = (n + l)‘+i - 1 - n.

k=l

5.7. Let p(p(x)) = p(x)“. Comparing degrees of the two sides yields k2 =
km, from which k = -00, 0 or m. If p(x) is constant, then it must be 0, 1
or, in the case of odd m, -1. If p(x) is nonconstant, it assumes infinitely
many values, so that p(t) -t” h as infinitely many zeros t = p(x). But then
p(t) = tm.

5.8. The equation is satisfied if p(t) = c, a constant polynomial, and q(t)
is any polynomial with q(0) = c. Suppose degp(t) = n > 1, degq(t) = r.
Then n = r(n - 1) f rom which n = r = 2. Hence, there exist coefficients a,
b, c, u, v, w for which

at2 + bt + c = u(2ut + b)2 + v(2at + b) + w.

Expanding and comparing coefficients yields 1 = 4uu, b = b + 2uv, c =
ub2 + vb + w, whence v = 0. Hence p(t) can be an arbitrary quadratic
at2 + bt + c while 4aq(t) = t2 - (b2 - 4uc).

5.9. The only constant polynomial is 0. Let p(t) = at + b. Then the identity
is satisfied as long as p(u) = a, so that p(t) = u[t + (1 - u)]. Let p(t) =
at2 + bt + c. By substitution, we arrive at the requirements a = l/2, b = 0,
whence p(t) = (1/2)t2 + c.
If degp(t) = 3, comparing the leading coefficients of both sides of

p’(p(t)) = p(p’(t)) (1)

yields p(t) = ( 1/9)t3 +.. e. However, obtaining the other coefficients by
substituting into (1) is an unappetizing task, and another tack is desirable.
Differentiate (1) repeatedly to obtain

p”(p(t))p’(t) = p’(p’(t))p”(t) (2)

P”‘(P(QP’W2 + P”(P(t))P”(t> = P”(P’(t))P”(t>2 + p’(p’(t))p”‘(t). (3)
If 0 is the sole zero ofp’(k), then p’(t) = t2/3 and we must have that p(t) =
t3/9. Otherwise, let r # 0 and p’(r) = 0. Then, from (2), p’(O)p”(r) = 0.
Either p’(0) = 0 or p”(r) = 0. In the latter case, since p”‘(t) is a nonzero
constant, (3) gives p’(0) = 0. Hence

p’(t) = t2/3 - rt/3