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Solutions to Problems; Chapter (^3 287)
b2 = 4(c+ 1). Th us, c+ 1 is a square, say u2, from which b = 2u. Conversely,
ifb=2uandc= u2 - 1, both t2 + bt + c and t2 + bt + c + 1 are evidently
reducible over Z.
(b) If a = 3, we are led to b2 = 4(4 + 3~). Choosing c = 7 makes 4 + 3c
a square, and we obtain the examples
3t2 + lot + 7 = (3t + 7)(t + 1)
3t2 + lot + 8 = (3t + 4)(t + 2).
7.14. Modulo xq - x - 1, it can be shown by induction that
xqpGx+r, (r=1,2 ,... ).
This is clearly true for r = 1. If it is true for r = k, then
Xq k+l z (x + 1) q Ir - = xQk + 1 E (x + k) + 1 = x + (k + 1).
Hence,
xm - l- x. XP..... x9p-1 -1~x(x+l)(x+2)...(x+p-1)--l
= - xp-x- 1.
If n = 1, then p = q and xp - x- 1 E 0. If n > 1, then q > p and
XP-x-1$0.
7.15. (x4 - 1)4 - x - 1 = (x4 - 1)4 - x4 + (x4 - 2 - 1) = (x’ - x - 1)
[(x4 - 1)3 + (x4 - 1)2x + (x4 - 1)x2 + x3 + 1). The second factor vanishes
for x = -1 and x = 0, and so it is equal to
x(x + l)(xlO - x9 + x8 - 3x6 + 3x5 - 2x4 + 3x2 - 2x + 1).
7.16. Let w be an imaginary cube root of unity so that w2 + w + 1 = 0.
Then, when t = w,
(t + 1)” - t” - 1 = [(w + 1)” - 1 - w”] = -(WZn + 1+ w”) = 0
when n E 1, 5 (mod 6). Hence t -w divides the given polynomial. Since
t2 + t + 1 is the product of two factors of the type t-w, this too divides the
given polynomial. The derivative of (t + 1)” -t” - 1 is n[(t + 1),-l - tnel].
When n E 1 (mod S), this too vanishes for t = w and hence is divisible by
t2 + t + 1. Prom these facts, the results can be obtained by setting x = ty.
7.17. The given polynomial can be written as a rational function with
numerator
(t4 - l)k+’ + (t + l)k+V-’