374 Answers to Exercises and Solutions to Problems
(c) First solution. The equation is equivalent to
8(z2 + XY + y2 + 1) = (x + y)(x2 + y2).
Prom this, it is clear that x and y have the same parity and that x + y > 0.
If x and y are both positive, then (x, y) # (1,l) and
x2 + y2 < x2 + xy + y2 + 1 < 2(x2 + y2)
so that 8 < x+y < 16. Trying out in turn y = lo-x, y = 12-x, y = 14-x
yields only the solutions (x, y) = (2, S), (8, 2).
If x and y have opposite sign, then
x2 + xy + y2 + 1 < x2 + y2,
sothatx+y<8.Tryingoutinturny=2-x,y=4-x,y=6-xyields
no further solutions.
Second solution. Since x and y have the same parity, let 2u = x + y and
v = xy. Then x2 + y2 = 4u2 - 2v and the equation becomes
8(4u2 - v + 1) = 2u(4u2 - 2v)
a (u - 2)” = 2u3 - 8u2 - 2 = 2(u - 2)(u2 - 226 - 4) - 18.
We have that u # 2 and
0 5 (x-y)2 = 4(u2- v)
= 4(u - 2)-‘[u3 - 2u2 - 2u3 + 8u2 + 21
= 4(u - 2)-l[-u2@ - 6) + 21 = 4(2 - u)-‘[u2(u - 6) - 21.
Hence, u - 2 ] 18 and 3 5 u 5 6. The only possibility is that (u, v) = (5,16),
so that (x, y) = (2,8), (8, 2).
- All the zeros must be negative, since f(x) > 0 for x > 0. Denote them
by -ri (1 5 i 5 n). Since rlr2 1.1 r, = 1, the AGM inequality yields
whence f(x) 2 (x + 1)” for x > 0. The result follows.
- Adding the equations yields (x+Y+z)~ = (x+y+z), whence x+y+z = 0
or x + y + z = 1. Prom the difference of the last two equations, we find that
(y-r)(y+z-2x+l)=O,whencey=zory+z=2x-l.Therearefour
cases:
(i) 2 + y + z = 0; y = z. then -2y = x = x2 + 2y2 = 6y2, so y =^0 or
y = -l/3. Hence, (x,y, z) = (O,O,O) or (2/3, -l/3, -l/3).
(ii) x+y+z = 0; y+z = 2x - 1. Then x = l/3 and we find that y and z
are the zeros of the quadratic 9t2+3t+l. Hence, (x, y, z) = (1/3,w/3,w2/3)
or (l/3, w2/3, w/3).
