Solutions to Problems; Chapter (^4 315)
whence ux = (x-y)(x-z), by = (y-x)(y-z), cz = (z-x)(z-y). Hence
ax(z - y) = by(x - z) a z(ax + by) = (u + b)xy
- bx-’ + ay-’ - (a + b)z-’ = 0.
Similarly
cx -’ - (u + c)y-l + az-’ = 0
-(b + c)x-’ + cy-’ + bz-’ = 0.
Prom any two of these equations, x : y : z = 1 : 1 : 1, SO that by + cz =
cz + ux = ax + by = 0, which is feasible only if a = b = c = 0.
8.20. If a = b, it is straightforward to verify that the equation is satisfied
iff x = y. Suppose a # b and let x = (1+ u)/(l - u), y = (1+ v)/(l - v).
The equations become
(ub + 1)(u2 + 1) = (a2 + l)(uv + 1)
(ub + l)(v2 + 1) = (b2 + l)(uv + 1).
(At this point, some obvious solutions can be noted: (u, v) = (a, b), (-a, -b),
(i, i), (-i, -i).) These equations lead to
(b2 + 1)(u2 + 1) = (a” + 1)(v2 + 1)
or
and
(u” + l)v2 = (b2 + 1)u2 + (b + u)(b - a)
Hence
(u2 + 1)v = (ab + 1)u + u(b - a)u-‘.
[(u” + l)(b2 + 1) - (ub + 1)2]u2 + [(a2 + l)(b + a)(b - a) - 2(ab + l)u(b - u)]
- a2(b - u)~u-~ = 0
- 0 = u4 + (1 - a2)u2 - a2 = (u2 - u2)(u2 + 1).
Therefore, the four obvious solutions are the only solutions and we can
now determine x and y.
8.21. First solution. Let u = x + y, v = xy. The three equations become
-3uv+z~=8,-2v+,z~=22,~~+uz+v=O.Hence(u+~)~=8
and (u + z)” + 22 2 = 22, whence (u + t,t2) = (2,9), (2w,^11 - 2w2),
(2w2, 11 - 2~). This leads to the solutions (x, y,z) = (-3,2,3), (2, -3,3),
(3,2, -3), (2,3, -3) and a number of nonreal solutions such as (2w,
-&l - 2w2, &l - 2~~). Here, w is an imaginary cube root of unity.
Second solution. The last equation is (z + x)(z + y) = 0. Let z = -2;
the case z = -y can be similarly handled. Then y3 = 8, so that y = 2, 2w,
2w2. Then 2x2 = 22 - y2 permits x to be determined.