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Solutions to Problems; Chapter 3 289

(Q - b)z] = (2~ - b)(02 - ab + b2). If a2 + b2 = ab (i.e. a/b is a primitive 6th
root of unity), the left side of the equation is undefined and the equation is
not valid. Hence, the equation is valid as long as b # 2a and a2 + b2 # ab.

8.3. An obvious solution is x = 0. Since, for every nonzero integer x,

x cannot be even. If x is odd, we must have that

which reduces to x2 - 2x -^3 = 0. The only other solutions are 2: = -1, 3.

8.4. First solution. Suppose z is a common zero of the two polynomials.
Then

z + 1 = z5 = z(az + b)2 = (a” - 3a2b + b2)z + (a3b - 2ab’)

Since I is nonrational (why?),

a4-3a2b+b2 = 1

a3b - 2ab2 = 1.

Eliminate b2 from (1) and (2) to get

5a3b=2a5-20-l.

Squaring (3) an d using (2) to eliminate b2 gives

25a5(a3b - 1) = 8a1’ - 16a6 - 8a5 + 8a2 + 8a + 2.

Use (3) to eliminate b and obtain

Jo + 3a6 - llc? - 402 - 4a - 1 = 0.

(1)

(2)

(3)

(4

(5)

But (5) has no rational roots. Hence the assumption of a common zero
leads to a contradiction.
Second solution. If z is a common zero, then z is a zero of the remainder
(a” - 3a2b + b2 - 1)~ - (a3b - 2ab2 - 1) when the quintic is divided by the
quadratic. This leads to (l), (2), which when solved as a linear system for
b and b2 yields

5a3b = 2a5 - 2a - 1 5ab2 = a5 - a - 3.

Elimination of b yields (5) and the argument proceeds as before.