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Solutions to Problems; Chapter 1 253

= sin2(A + B) + (q - 1) tan(A + B) sin(A + B) cos(A + B) + q cos2(A + B)

= sin2(A + B) + qsin2(A + B) - sin2(A + B) + q cos2(A + B) = q.

8.2.

n

x(x + i - l)(z + i) = pn * nx2 + n2x + (n+lbw)-pn=O

i=l^3

Hence 2r + 1 = -n and r(r + 1) = (n2 - 1)/3 - p. Eliminating r yields
n2=12p+1.1fp=10 therm= 11 and the solutions of the quadratic are
-5 and -6. In general,’ n must have the form 6k f 1, whence p = 3k2 f k.
8.3. Solution 1. Let the roots of ax2 + bx + c = 0 be P and s. Then one is
the square of the other if and only if

0 = (r - s2)(s - r2) = rs + r2s2 - [(r + s)3 - 3rs(r + s)],

which is equivalent (using r + s = -b/u, rs = c/a) to

0 = u2c + ac2 + b3 - Sabc.

Solution 2. Suppose that the roots of ax2 + bx + c = 0 are u and u2.
Then u + u2 = -b/a, whence au2 + au + b = 0. But au2 + bu + c = 0,
so (b- a)~ = (b - c). If a = b, then b = c and the equation is a constant
multiple of x2 +x + 1 = 0. If a # b, then substitute u = (b - c)/(b - a) into
au2 + bu + c = 0. In either case, we obtain

b3 + 02c + ac2 - 3abc = 0.

On the other hand, let u and v be the roots of ax2 + bx + c = 0, and
suppose that b3 + a2c + ac2 - 3abc = 0. Then b = -a(u + v), c = auv leads
to 0 = (u - v2)(v - u2), so one of u and v is the square of the other.
Solution 3. If u and u2 are the roots of the given equation, then u is a
common root of the two equations ax2 + bz + c = 0 and az2 + ax + b = 0
(since U+ u 2 = -b/a). Hence (b - a)u = (b - c). Substituting this into the
equation yields the condition a(b - c)~ + b(b - a)(b - c) + c(b - a)2 = 0.
On the other hand, suppose the condition holds. If a = b, then b = c
and the equation is a constant multiple of x2 + z + 1 = 0, each of whose
roots is the square of the other. If a # 6, then u = (b - a)-‘(b - c) satisfies
ax2 + bx + c = 0 and ax2 + a,z + b = 0. Hence u + u2 = -b/a and u2 must
be the second root of ax2 + bx + c = 0.

8.4. Let q(t) = p(n + t) = t2 + bt + c. Then p(n)p(n + 1) = c(1 + b + c) =
q(c) = p(n + c). Also, p(n)p(n + 1) = p(n - c - b).

8.5. The discriminant of the second quadratic is

4(” + lg2 - 12(q + ap) = (2a - p)2 + 3(p2 - 4q),