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# Eisenstein's Proof

To set the stage, we recall a few consequences of the fact that the residue classes modulo a prime p form a field . 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).

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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.)

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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,]
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Next: Eisenstein versus Gauss Up: Eisenstein's Misunderstood Geometric Proof Previous: Introduction

D. Pengelley and R. Laubenbacher

Tue Feb 9 17:35:23 MST 1999