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290 Answers to Exercises and Solutions to Problems

8.5. First solution. Let U, v, w be the zeros of x3 + ax2 + 11x + 6 and u,

v, % the zeros of z3 + bx2 + 14x + 8. Then

u+v+w=--a uv+uw+vw=11 uwv=-6

u+v+z=-b uv + U% + V% = 14 uvz = -8

* w(u + v) = 11- uv z(u + v) = 14 - uv 6% = 8w

a 6(14 - uv) = 8(11- uv) + uv = 2

* w=-3, %=-4*u+v=-3.

Hence a = 6, b = 7.

Second solution. The common zeros of two polynomials are zeros of their

difference (Q - b)x2 - 3x - 2. Now

6 + 11x + .x2 + x3 = (2 + 3x + (b - a)x2)(3 + x)

+ (4a - 3b - 3)x2 + (1 + a - b)x3.

Since 2+3x+(b-a)x2 divides 6+11x+ax2+x3, it follows that 4a-3b-3 =

l+a-b=O,ore=6,b=7.

8.6. Since the r-ai are distinct nonzero integers and at most two can have

the same absolute value, their absolute values arranged in increasing order

are respectively at least equal to 1, 1, 2, 2, 3, 3,... , n, n. Hence

However, since ll(r - oi) = (-1)“(n!)2, equality actually occurs. Thus,

the numbers r - ai are +l, -1, +2, -2,... ,+n, -n in some order and

r-al+r-a2+.. .+r-az,=l-1+2-2+...+n-n=O.Theresult

follows.

8.7. Let t3 - mt2 - mt - (m2 + 1) have an integer zero t. Then t must be

such that the quadratic equation

m2 + (t2 + t)m - (t3 - 1) = 0

has an integer solution m. The discriminant of this quadratic is

t4 + 6t3 + t2 - 4 = (t2 + 3t - 4)2 + (24t - 20) = (t + 4)2(t - 1)2 + 4(6t - 5).

If t 2 1, then

(t2 + 3t - 4)2 + (24t - 20) > (P + 3t - 3)2,

so that t 5 8. If t 5 -5, then

(t2 + 3t - 4)2 + (24t - 20) 5 (P + 3t - 5)2,