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Answers to Exercises; Chapter 3 273

1.8. Assume that a > 0 and note that

at2 + bt + c = u[(t - b/2u)2 - (1/4e)(b2 - 4ac)].

Apply Exercises 7 and 6. The quadratic at2 + bt + c is

(i) always reducible over C

(ii) reducible over R iff b2 - 4ac 1 0
(iii) reducible over Q iff b2 - 4ac is a perfect square.
Reducibility over Z requires more careful analysis. It is necessary that
the discriminant b2 - 4ac be a perfect square, say d2. On the other hand,
if this is true, then

at2 + bt + c = (1/4a)(2at + b - d)(2ut + b + d) = (u/v)(pt + q)(rt + s)

where gcd(u, v) = gcd(p, q) = gcd(r, s) = 1. Comparing coefficients of both
sides, we find that v must divide pr, ps + qr, qs. Suppose d is a prime
divisor of v. Then, if (say) dip, then d)(q. Hence dls and d$r. But then
dips and d$qr, so that d$ps + qr, a contradiction. Hence v has no prime
divisors, so v = 1. Therefore at2 + bt + c is reducible over Z if b2 - 4uc is a
perfect square.

1.9. (5/4)t2 - (31/45)t - (8/5) = ((5/6)t - 6/5)((3/2)t +4/3), for example.
1.10. (a) Express the coefficients off over a least common divisor d. Then
let c be the greatest common divisor of the numerators.
(c) We can write (c/d)g(t) = f(t) = fl(t)fz(t) where fr and f2 have
rational coefficients. Each fi can be written in the form (ci/di)gi, as in (a).
The result follows with a/b = clczd/dldzc.
(d) Let gl(t) = Cu&, gz(t) = Cvjtj. Let p, r, s be as specified. The
coefficient of tr+’ in the product gl(t)gz(t) is

c uiv,+,-i + u,v, + c uivr+a-i.

i<r i>r

Since the prime p divides ui for i < r and Vr+s-i for i > r, p divides both
sums. But p.jwrvb, so that the coefficient of tr+’ in (a/b)gl(t)g2(t) is not
an integer, yielding a contradiction.

1.11. Let f(t) = t” + anvltn-l +... + alt + ac, and suppose that r = u/v
(in lowest terms). Then ‘1~” + ‘&r aiunsivi = 0. If a prime p divides v,
then p must also divide U, contradicting the coprimality of u and v.

1.12. Let h(t) = f(t)g(t), where f(t) = Gait’, g(t) = Cbjtj. Then

co = aobo

cl = aobl + albo

c2 = aobz + ulbl + uzbo