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,]\