Answers to Exercises; Chapter 48.30. We have the following table of solutions:Modulo the congruence is satisfied by n congruent to29733 = 27 1 (mod 3)
34 =81 1,4 (mod 9)35 = (^243 19) (mod 27)
36 = 729 46 (mod 81)
37 = 2187 208 (mod243)
Suppose m 1 5 and n - a (mod 3”-2) satisfies f(n) E 0 (mod 3”).
Then a E 1 (mod 9) and f(a) = 3”k. Let n = a + 3”-2b, so that n2 G
a2 + 3”-2 2ab and n3 G a3 + 3”-la2b (mod 3”+‘); then
f(n) E f(a) + 3”-‘a2b + 3”-’ e264ab - 3”-’ .37b (mod 3”+l).
Hence f(n) E 0 (mod 3”+‘) e 3k + (a2 + 264a - 37)b E 0 (mod 9)
3k + (a - 4)(a - 2)b E 0 (mod 9). Since a E 1 (mod 9), a - 4 = 3u,
where u is not divisible by 3, so that
f(n) E 0 (mod 3”+‘) k + u(a - 2)b E 0 (mod 3).
The last congruence is uniquely solvable for b modulo 3.
Since f(n) z 0 (mod 3”) h as a unique solution a modulo 3”-2 (and
therefore nine solutions a + ~3”~~) where 0 5 c 5 8, modulo 3”), it follows
that f(n) E 0 (mod 3”+‘) has a unique solution a + 3”-2b modulo 3”-‘,
and the required result follows by induction.
8.31. Suppose such a representation were possible. Then g(t) cannot be
constant and tg(t)f (t + 1) - tg(t + l)f(t) = g(t)g(t + 1). Since f(t) and
g(t) have no common divisor of positive degree, g(t) I tg(t + l), so that
tg(t + 1) = u(t)s(t). BY considering degrees and leading coefficients, and
by noting that g(t) is not constant, we see that u(t) = t - c where c # 0.
Hence g(t) has zeros c + k for k = 1,2,.. ., which contradicts g(t) having
positive degree.
Answers to Exercises
Chapter 4
1.1. (x,y,z) = (7,9,13).
1.3. (x,y, z) = (10,4,-S), (-10, -4,6).
1.4. If (2, y, Z) satisfies
ax + by + cz = px + qy + rz = 0, (*>