The Quadratic Reciprocity Theorem compares the quadratic character of two primes with respect to each other.

**Quadratic Reciprocity Theorem*** If p and q are
distinct odd primes, then*

Here is Eisenstein's proof, closely following both his own language and notation (which he conveniently and successfully abuses).

*\*

Consider the set . Let
**r** denote the remainder
of an arbitrary multiple **qa**. Then it is apparent that the list of
numbers agrees with the
list of numbers **a**, up to multiples of **p**. (For clearly each*
*of the numbers has
even least positive residue, and if there were duplication among these
residues, e.g.

then . Since the
**a**'s are distinct, it follows that ,
which cannot occur since
and is even.) Thus

from which it follows that . Recalling Euler's Criterion that , this produces

so one may focus solely on the parity of this exponent. Clearly

where is the greatest
integer function. Since the elements **a** are all even, and **p**
is odd, it follows that
and thus

(Here Eisenstein remarks that since up to this point **q**
need not have been an odd prime, but merely a number relatively prime to
**p**, one can easily obtain the Ergänzungssatz
from the above formula. This we leave as an exercise for the reader.)

\

Eisenstein now uses a geometric representation of the exponent in this
last equation to transform it twice while retaining its parity: This exponent
is precisely the number of integer lattice points with even abscissas lying
in the interior of triangle **ABD** in the Figure (note that no lattice
points lie on the line **AB**).

Consider an even abscissa .
Since the number of lattice points on each abscissa in the interior of
rectangle **ADBF** is even, the number of
lattice points on the abscissa below **AB** has the same parity as the
number of lattice points above **AB**. This in turn is the same as the
number of points lying below **AB** on the odd abscissa **p-a**.
This one-to-one correspondence between even abscissas in triangle **BHJ**
and odd abscissas in **AHK** now implies that
where is the number of
points inside triangle **AHK**, and thus

Reversing the roles of **p** and **q** yields
where is the number of
points inside triangle
Since the total number of points inside both triangles is simply ,
one may now conclude that

Even the normally modest Eisenstein could not restrain his pleasure with this proof:

``How lucky good Euler would have considered himself, had he possessed these lines about seventy years ago.'' [7, p. 174,]\