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Solutions to Problems; Chapter (^8 371)
Second solution. Let y = z + u. Then
(3u + 2)x2 + (3u2 + 2u)x + (u” + 8) = 0.
The discriminant of this quadratic equation is
-3u4 + 4u3 + 4u2 - 962s - 64 = -(3u + 2)(u3 - 2u2 + 32)
which is a positive square only for u = -2. This leads to the solutions
(xc, Y) = (0, -a (2,O).

9. Since fi is nonrational and x, y, z, t are to be rational, the equation is
   equivalent to the system

x2 + 2y2 + z2 + 2t2 = 27

xy + zt = 5.

Hence

(5x2 - 27xy + 10~~) + (5 .z2 - 27zt + 10t2) = 0

* (5x - 2y)(x - 5y) + (5z - 2t)(z - 5t) = 0.

Trying x - 5y = z - 5t = 0 reduces both equations of the system to

y2 + t2 = 1 and we obtain, for example, the solution

(x,y,z,J)=

(

-2% 2s 5 1-

1+s2’1+s2’ 1+s2 ‘1+s2 >

where s is any rational. [Trying 5x- 2y = z-St = 0 reduces both equations

to 2y2 + 25t2 = 25. If we write y = 5u, then this becomes 2u2 + t2 = 1

and we obtain, for example, the solution (x, y, z, t) = (4/3, 10/3, 5/3, l/3).

Trying 5x - 2y = 5z - 2t = 0 reduces both equations to 2(y2 + t2) = 25

and we obtain for example the solution (I, y, z, t) = (1, 5/2, 1, 5/2).]

10. Let u = VT& and v = vr&. Observe that

(1) u+v=z.

(2) u3 + 213 + 3uv(u + v) = 23 * 3 2121% = z3 - u3 - v3 j 2121 is a rational

number. Suppose uv = p/q in lowest terms.

(3) u3v3 = (x + &)(z - fi) = x2 - y, an integer. Then p3 = (x2 - y)q3,

so that any prime which divides q must divide p. Since gcd(p, q) = 1,

we must have q = 1. Hence uv = p, an integer.

Thus, u and v are zeros of the quadratic t2 - zt +p, with the result that
2u = z + fi and 2v = z - 6, where w = z2 - 4~. The value of w can
be either positive or negative. If positive, u and v are real and we decide
arbitrarily that u 2 v. If w is negative, then fi is consistently taken to
be one of the square roots of 20.