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288 Answers to Exercises and Solutions to Problems

and denominator (t + l)k. It suffices to show that the numerator is divisible
by t5 + 1, or, equivalently, vanishes when we set t5 = -1. This can be seen
by expanding the numerator binomially and pairing the terms:

t4k+4 -( k;1)t4k+( ‘;‘)+4-( k;l)t-+...+(-qk+l

• t4k-1 +( “:‘)t4k+( k;1)t4k+l+( ‘;‘)4”f2+...+tS’.

7.18. Let p be any prime exceeding 3. We show that x2 - x + 1 divides
x2P - xr + 1. Let w be a zero of the quadratic. Then w is a primitive 6th
root of unity. Since p - 1 or 5 (mod 6), Wp is also a primitive 6th root of
unity, so that w is a zero of x2P - xp + 1, from which the divisibility follows.
Suppose that m = kp, with p as above’ and x = 2k. Then it follows that
4k-2k+1divides4m-2m+1.

7.19. n is clearly odd. Let n + 1 = 2r. Then

4" + n4 = (2" + n2)2 - n22”+l = (2” + n2’ + n2)(2” - n2’ + n”).

If r = 2 or 3, the number is easily cnecked to be composite. If r 2 4, then
2n + n2 > 2” = 2’-‘2’ > (2r - 1)2’, so that both factors of the number
exceed 1 and the number is composite. If r = n = 1, the number is prime.

7.20. Let 2k= m + 1. Then

22m + 1 = (2” + 1)2 - 22k = (2” + 2” + 1)(2m - 2” + 1).

There are four cases to be considered. If (k, m) E (2,3) or (3,1) (mod 4),
then 2m - 2k + 1 E 0 ( mod 5); if (k, m) E (1,1) or (0,3) (mod 4), then
2" + 2k + 1 E 0 (mod 5). I n any case, one of the factors is divisible by^5
and, when m > 3, it is straightforward to check that both factors exceed 5.

7.21. f(x, Y) = (X - Yhdx, Y) + fb49 = (Y - +a 4 =b dx, Y) =
-q(y, x). Since q(x, y) is a polynomial, this last equation persists for y = x,
so that q(x, x) = 0. By the Factor Theorem, q(x, y) = (x - y)r(x, y) and
the result follows.

7.22. The equation can be rewritten

0 = nx”-l + (n - 1)~“~~ +... + 2x - (n - l)(n + 1)

= [nx - (n + l)][~“-~ + 2xns3 + 3xnB4 +... + (n - l)],

whence x = (n + 1)/n is a solution.

8.1.. yx = x2 + 1 j x2 - x3 - x = x2 - yx2 = (1 - y)x2 = y2x2 =
x4 + 2x2 + 1 j x4 + x3 + x2 + x + 1 = 0, whence the result follows.

8.2. If b = 2u, both sides of the equation are undefined. Let b # 2~. Now
a3 + b3 = (a + b)( u2-ub+b2) anda3+(u-b)3=(2a-b)[a2-u(a-b)+