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Solutions to Problems; Chapter 4 311

Checking the possibilities leads to (x, y) = (2,2), (-2, -2), (3, l), (-3, -l),
and four other solutions given by 3x2 + 2 = 0, x + y = 0 and 17y2 + 4 = 0,
x+3y=o.
8.7. Let u = -x+y+x, v = x-y+%, w = x+y-Z. The equations become
VW = u2, uw = b2, uv = c2. Then b2v = a2u, w = c2 lead to u = fbca-l,
v=fucb-1.Alsow=fabc-1.Then2x=v+w,2y=u+w,2z=u+v.
8.8. From the first two equations x : y : z = (b - c) : (c-a) : (u - b). Using
this in the last equation leads to (x, y, Z) = (b - c, c - a, a - b).

8.9.Sincexy-xr=b-c,wefindthat2xy=u+b-cand2x%=u-b+c.

Hence %(a + b - c) = y(u - b + c). This, with 2yz = -a + b + c yields

2(u + b - c)z2 = (u - b + c)(-a + b + c).

Similarly, x2 and y2 can be found.

8.10. (a) w = u + v - u-l, w-l = u + v - v-l

* l=(u+v)z-(u+v)(u-l+v-l)+u-lv-l

j 1 - u-%--l = (u + v)2(1- u-%-1)

3- uv=l,u+v=l or u+v=-1.

Hence (u,v, w) = (z, z-l,%), (z, 1 - Z, (Z - 1)~~‘) or (z, -(l + z),

-(l + z)z-I), for arbitrary nonzero z.

(b) Equating first and second, first and third, second and fourth, third

and fourth members, respectively, yields

uv(w - l)(w - u) = w(v - l)(W - 1)... (1)

UV(W - l)(vw - 1) = w(u - l)(uv - 1)... (2)

u(uv - l)(vw - 1) = vw(w - l)(u - 1)... (3)

vw(uv - l)(w - u) = u(v - l)(w - 1)... (4)

The solutions in which any one of u = 1, v = 1, w = 1, 21 = w, uv = 1,

VW = 1 occur are covered by the following: (21, v, w) = (1, Z, z-l), (z, 1, z),

( ZJ -I, 1) where z is nonzero.

If none of the foregoing possibilities occur, then we can cancel freely in

manipulating the above four equations. From (1) and (2), we have

(u - l)(w - u) = (v - l)(vw - l),

and from (3) and (4),

(vw)2(u - l)(w - u) = 212(v - l)(vw - 1).

Hence, u = VW or u = -VW.