Prime Numbers

2.4 Exercises 111
2.27. Using Theorem 2.3.7 prove the two equalities in relations (2.12).
2.28. Here we prove the celebrated quadratic reciprocity relation (2.11) for
two distinct odd primes p, q. Starting with Definition 2.3.6, show that G is
multiplicative; that is, if gcd(m, n) = 1, then
G(m; n)G(n; m) =G(1; mn).
(Hint: mj 2 /n + nk 2 /m is similar—in a specific sense—to (mj + nk) 2 /(mn).)
Infer from this and Theorem 2.3.7 the relation (now for primes p, q)
p q
=(−1)
q p
(p−1)(q−1)/4 .
These are examples par excellence of the potential power of exponential
sums; in fact, this approach is one of the more efficient ways to arrive at
reciprocity. Extend the result to obtain the formula of Theorem 2.3.4 for
2
p .
Can this approach be extended to the more general reciprocity statement (i.e.,
for coprime m, n) in Theorem 2.3.4? Incidentally, Gauss sums for nonprime
arguments m, n can be evaluated in closed form, using the techniques of
Exercise 1.66 or the methods summarized in references such as [Graham and
Kolesnik 1991].
2.29. This exercise is designed for honing one’s skills in manipulating Gauss
sums. The task is to count, among quadratic residues modulo a prime p, the
exact number of arithmetic progressions of given length. The formal count of
length-3 progressions is taken to be
A(p) =# (r, s, t) :
r s t
p = p = p =1;r = s; s − r ≡ t − s (mod p) .
Note we are taking 0 ≤ r, s, t ≤ p − 1, we are ignoring trivial progressions
(r, r, r), and that 0 is not a quadratic residue. So the prime p = 11, for which
the quadratic residues are {1, 3, 4, 5, 9}, enjoys a total of A(11) = 10 arithmetic
progressions of length three. (One of these is 4, 9, 3; i.e., we allow wraparound
(mod 11); and also, descenders such as 5, 4, 3 are allowed.)
First, prove that
p − 1 1
A(p) =− +
2 p
p−1
e 2πik(r−2s+t)/p ,
k=0 r,s,t
where each of r, s, t runs through the quadratic residues. Then, use relations
(2.12) to prove that
p − 1
2 −1
A(p) = p − 6 − 2 − .
8
p p
Finally, derive for the exact progression count the attractive expression
p − 2
A(p) =(p− 1) .
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