Exercises 39
(b) For some real number x0 and constants c1,c2 0, we have
(1.18) c1x #{a A : a x} c2x for all x x0.
Prove that P is infinite, and that in fact

19. (Continuation)
(a) For each nonzero Gaussian integer α put α =
Show that

π −1 diverges, where the sum is over all Gaussian primes π.
Deduce that

p≡1 (mod 4)
diverges, where the sum is over ra-
tional primes p 1 mod 4.
(b) For each nonzero polynomial F (T) Fq[T], put F :=
qdeg F
Show that

diverges, where P ranges over the irreducible
elements of Fq[T].
20. This exercise outlines a proof of Theorem 1.21 via algebraic number
theory. Let m be a positive integer, and let ζ be a primitive mth root
of unity. Put K = Q(ζm), and identify Gal(K/Q) with
H be a subgroup of
and let L K be the fixed field of H.
(a) Say that two sets of rational primes P1 and P2 eventually coincide
if their symmetric difference is finite; in this case we write P1
P2. Prove that P1
= P2, where P1 is the set of primes for which
p mod m H and P2 is the set of primes which split completely in
L. Hint: If p is a prime not dividing m, analyze how the Frobenius
element of p in Gal(K/Q) behaves upon restriction to L.
(b) Let θ be an algebraic integer for which L = Q(θ). Let F be the min-
imal polynomial of θ. Prove that P2, and hence also P1, eventually
coincides with the set of prime divisors of F . Hint: L/Q is Galois,
so an unramified rational prime splits completely in L exactly when
it has a degree 1 prime factor; now apply the Kummer-Dedekind
21. (P´ olya [P´ ol21]; see also [MS00]) Suppose that a and b are nonzero in-
tegers and a = ±1. Let P be the set of primes for which the exponential
b (mod p) has a positive integer solution k. In other
words, P is the set of primes which divide some term of the sequence
a b,
b, . . . .
This exercise outlines a proof that P is always an infinite set.
We may suppose that b is not a power of a, as otherwise P contains
every prime. We assume for the sake of contradiction that P is finite.
(a) For each p P and each k 1, define integers vp,k 0 by writing
b = ±
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