Unknown

286 Answers to Exercises and Solutions to Problems

7.10. Let u = r + s, v = rs. Then q(x) = x2 - ux + v and p(x) =

(x - r)(x -s) + (r + s - x) = x2 - (u + 1)x + (u + v). Expressing the given

polynomial as the product of p(x) and q(x) and comparing coefficients leads

to

2a+l=-(2u+l)*a+u+l=O

(a - 1)2 = u2 + 2(u + v) * 2(u + v) = (a - 1)2 - u2 = -4a

4 = v(u + v) = -2av.

Since u = -a - 1 and u + v = -2~2, it follows that v = --a + 1 and that

4 = 2a2 - 2a. Hence a = 2 or a = -1.

Since b = -u(u+v)-v(u+l) = -(3u2+u), we have that (a, b) = (2, -14)

or (-1, -2). Thus, the possibilities are

x4 + 5x3 + x2 - 14x + 4 = (x2 + 3x - 1)(x2 + 2x - 4)

x4-x3+4x2-2x+4=(x2+2)(x2-x+2).

7.11. 320n2+ 144nl243 = (an+ b)(cn+d) implies that b and d are powers

of 3, gcd(uc, 3) = 1 and ad + bc = 144 z 0 (mod 9). Hence, without loss of

generality, we can take b = 9, c = -27. We find that 320n2 + 144n - 243 =

(8n + 9)(40n - 27).

Hence

3(81”+l) + (16n - 54)9”+’ - (320n2 + 144n - 243)

= [3(9”+‘) + (40n - 27)][9”+l - (8n + 9)]

= [27(9” - 1) + 40n][9(9” - 1) - 8n]

= 82[27(9”-1+9n-2+9”-3+.. .+1)+5n][9(9”-1+9”-2+9”-3+.. .+1)-n].

It is straightforward to check that, modulo 8, for each n, each factor of the
last member in square brackets is congruent to 0. Hence the given quantity
is divisible by 84 = 212.

7.12. If n = 2m, a = 2b, then x” + xa + 1 = (x”’ + x’ + 1)“. If f(x) =
xn + x” + 1 with exactly one of n and a even, then f’(x) = xk where
k = n - 1 or a - 1, and gcd(f, f’) = 1. If n and a are both odd, then
f’(x) = x-1 + xa-‘, so that f(x) = xf’(x) + 1 and gcd(f, f’) = 1. Since
any repeated factor of f( x ) must also divide f’(x), the result follows.
7.13. If at2 + bt + c and at2 + bt + c + 1 are both reducible over Z, then
for some integers m and n,

b2 - 4ac = m2 and b2 - 4ac - 4u = n2.

Hence 4a = m2 - n2.
(a) In the case that a = 1, this yields 4 = m2 - n2 = (m - n)(m + n).
Since m and n have the same parity, we must have m = 2, n = 0, whence