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256 Answers to Exercises and Solutions to Problems

It4 - i] = 121 - i] = J4cos2 e + 2sin #I = Jz.

Hence u and v lie on a circle with center i and radius 4. The line joining

u and v passes through -i and makes an angle 8 = 4/2+ n/4 with the real

axis. As C#J moves from 0 to 7r, u traces that part of the circle in the upper

half plane from 1 to -1 while v traces that part in the lower half plane

from 1 to -1.

8.17. The equation can be rewritten (x-~)~+(x-a)(b+c)+b~+c~- bc =

0. Solving this for x - a by the quadratic formula yields the solutions
   a + bw + cw2 and a + bw2 + CU, where w is an imaginary cube root of 1.
   8.18. If ax2 + bx + c and ax2 + bx - c are both factorable over Z, then
   b2 - 4uc = q2 and b2 + 4uc = p2 for positive integers p and q, which must
   have the same parity. Let 2u = p + q and 2v = p - q. Then b2 = u2 + v2
   and 2ac = uv. One of u and v must be even, say v = 2w.
   Noting that (r + si)2 = u + 2wi implies that (r2 + s2)2 = u2 + 4w2, we
   solve r2 - s2 = u, rs = w for r and s. Thus, r2 and -s2 are the roots of
   the equation
   t2 - ut - w2 = 0
   whence r2 = (lb1 + u)/2 and s2 = (lb1 - u)/2. It remains to show that r and
   s are integers. Now, b and u have the same parity and satisfy

(.y) (q2)=w2.

Let d be a prime divisor of w. Then d divides exactly one of the factors
on the left. Otherwise, d would divide their sum b. But d divides UC = uw,
which contradicts the coprimality of b and UC. It follows that each factor
on the left is a square, and so r and s are integers. It is straightforward to
show that r and s are coprime.
Hence, Ibl = r2 + s2 and UC = rs(r2 - s2). On the other hand, if these
conditions are satisfied, then
b2 - 4uc = (r” - 2rs - s2)2

and
b2 + 40~ = (r2 + 2rs - s2)2
and the quadratic can be factored.
The condition on a, b, c cannot be weakened to require merely that their
greatest common divisor be 1 (as specified in the source of this problem).
For a discussion, see Amer. Math. Monthly 47 (1940), 187-188.

8.19.

acos20 + 2hcosBsinB+bsin20

= (1/2)[(~ + b) + (U - b) cos 28 + 2h sin 281

= (1/2)[(u + 6) + d(u - b)2 + 4h2 sin(28 + 4)]