.
Fermat's Little Theorem, that 
for any integer b not divisible by p, holds because the nonzero
residue classes form a (cyclic) group of order p-1 under multiplication.
When p is odd, the squaring map 
has kernel 
, so its image,
the squares or quadratic residues modulo p, form a subgroup
of order 
and the nonresidues
form its coset. The quadratic character of a residue class 
is specified by the Legendre symbol: 
if b is a quadratic residue mod p and 
if not. From 
, it follows
that 
for any 
.
But if 
, then 
,
so the quadratic residues are all roots of the polynomial 
.
Since this polynomial can have no more than 
roots in the field 
, we
conclude that its roots are exactly the quadratic residues. That is, we
have Euler's Criterion: 
for any integer b not divisible by p.
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,]\
