proofs

26 The law of quadratic reciprocity
Let p and q be odd primes, and consider ( q p
). Suppose iq is a multiple of
q that reduces to a negative residue r i < 0 in the Lemma of Gauss. This
means that there is a unique integer j such that − p 2
< iq − jp < 0. Note
that 0 < j < q 2 since 0 < i < p 2 . In other words, ( q p ) = (−1)s , where s is
the number of lattice points (x, y), that is, pairs of integers x, y satisfying
0 < py − qx < p 2 , 0 < x < p 2 , 0 < y < q 2 . (3)
Similarly, ( p q ) = (−1)t where t is the number of lattice points (x, y) with
0 < qx − py < q 2 , 0 < x < p 2 , 0 < y < q 2 . (4)
Now look at the rectangle with side lengths p 2 , q 2
, and draw the two lines
parallel to the diagonal py = qx, y = q p x+ 1 2 or py −qx = p 2 , respectively,
y = q p (x − 1 2 ) or qx − py = q 2 .
The figure shows the situation for p = 17, q = 11.
11
2
p = 17 q = 11
s = 5 t = 3
`q ´
= (−1) 5 = −1
p
`p ´
= (−1) 3 = −1
q
R
The proof is now quickly completed by the following three observations:
17
2
S
1. There are no lattice points on the diagonal and the two parallels. This
is so because py = qx would imply p | x, which cannot be. For the
parallels observe that py − qx is an integer while p 2 or q 2
are not.
2. The lattice points observing (3) are precisely the points in the upper
strip 0 < py − qx < p 2
, and those of (4) the points in the lower strip
0 < qx − py < q 2
. Hence the number of lattice points in the two strips
is s + t.
3. The outer regions R : py − qx > p 2 and S : qx − py > q 2
contain the
same number of points. To see this consider the map ϕ : R → S which
maps (x, y) to ( p+1
2
− x, q+1
2
− y) and check that ϕ is an involution.
Since the total number of lattice points in the rectangle is p−1
2
· q−1
2 , we
infer that s + t and p−1
2
· q−1
2
have the same parity, and so
(q
p
)(p
q
)
= (−1) s+t = (−1) p−1 q−1
2 2 . □