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Solutions to Problems; Chapter 4 317

() yields (u - 3)(3u3 + u2 - 4u - 12) = 0. C onversely, any solution of this
equation leads to a solution of x+y+% = 4, x2-22 = 24, x2+y2+z2 = 30,
and hence of the given system.
Case (ii): From x + % = -4 - y, x - % = 24(-4 - y)-‘, we obtain
(2, y, z) = (v + 6v-5 -4 - 2v, v - 6~l) where 2v = -4 - y. Substituting
this into () yields (v + 3)(3v3 - v2 - 4v + 12) = 0. Any solution of this
equation yields a solution of the original system.
Hence (x,y,%) = (5,-2, l), (-5,2,-l), (U + 621-l, 4 - 2u, u - 6u-l)
where u(u - 1)(3u + 4) = 12 or (v + 621-l, -4 - 2v, v - 6~~‘) where
v(v + 1)(3v - 4) = -12.
8.25. For any quadratic polynomial f(t), it is easy to verify that

Ckf(k2)xk = uf(u).

In particular, if f(t) = (t - a)2, we obtain that

Ck(k2 - a)2xk = 0

whence each term of the sum on the left must vanish. This implies either
that each Xl, vanishes, in which case a = 0, or that a = m2 for m equal to
one of 1, 2, 3, 4, 5, in which case x ,,, = m and the remaining zk vanish.

8.26. Suppose we have a real solution (x, y,%). Then x + y = 2 - % and
xy = (% - l)“, so that

0 5 (x - y)2 = (2 - %)” - 4(% - 1)2 = %(4 - 32).

Hence, we must have 0 5 % 5 4/3. Then p = xy% = z(z - 1)2 assumes
all values between 0 and 4/27 inclusive (sketch a graph of the function
Z(% - 1)2).
Conversely, if 0 5 p 5 4/27, then the equation z(z - 1)2 = p is solvable
for real % with 0 5 % 5 4/3. For such %, the quadratic t2 - (2- z)t + (% - 1)2
has real zeros x, y, and we have a real solution (x, y, %) of the given system.

8.27. The second equation implies that

x+Y X-Y

-+-

X-Y x+Y

=f 3+;

[ 1

whence one of x = f2y, y = f2x holds. The solutions (x, y) to the
;yat$ns are (2,1), (-2,-l), (1, 2), (-l,-2), (2i,-i), (-2i,i), (i,-2i),
2,. 2..

8.28. The first two equations represent spheres in Cartesian S-space with
radii 7 and centers (5,2,6) and (11,7,2). Since the centers are less than 14
units apart, the spheres intersect in a circle lying on the plane 12x + 1Oy -
82 = 109. Since the plane 38x - 56y - 13% = 0 passes through the point

(8,9/2,4) w h ic h is at the same time midway between the centers of the