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Eisenstein versus Gauss

Gauss himself considered his third proof to be the most direct and natural of his demonstrations. In introducing it he said:

``For a whole year this theorem tormented me and absorbed my greatest efforts until at last I obtained a proof ... . Later I came across three other proofs which were built on entirely different principles. I do not hesitate to say that until now a natural proof has not been produced. I leave it to the authorities to judge whether the following proof which I have recently been fortunate enough to discover deserves this description.'' [10, p. 113,]

While Eisenstein essentially follows the same outline as Gauss, each feature of his approach displays great clarity and insight, and offers an elegant view while shortening the path taken by Gauss.

Gauss' third proof begins with his Lemma, which says that


with obtained as follows: Let

Then is defined to be the number of least positive residues of the set which lie in .

Instead of using Gauss' Lemma, Eisenstein derives equation (1), with the algebraic expression in the exponent, which is then more easily converted into the key equation


common to both proofs, than is equation (3). While Eisenstein's algebraic exponent is easily transformed into the exponent in (4) via (2), Gauss must establish a number of technical properties of the greatest integer function and apply them to relate to the exponent in (4). Eisenstein's use of the set , as opposed to Gauss' , allows him to count the same elements as Gauss' Lemma, but via the expression , leading quickly to (4):

``The main difference between my argument and that of Gauss is that I do not divide the numbers less than p into those less than and those greater than , but rather into even and odd ones.'' [7]

Eisenstein now applies his two clever geometric transformations to convert the exponent into the number of lattice points in triangle AHK (mod 2). After doing the same for , yielding the number of lattice points in AHL, the proof is completed simply by counting the lattice points in rectangle AKHL. gif Gauss, on the other hand, in essence performs the same two transformations, and counting, without availing himself of the geometric presentation. He actually counts the lattice points using algebraic properties of the greatest integer function. This makes the remainder of his proof lengthy and nonintuitive, and forces him to consider separate cases depending on the congruence classes of p and q (mod 4). (For a more detailed comparison, see [8].)

next up previous
Next: References Up: Eisenstein's Misunderstood Geometric Proof Previous: Eisenstein's Proof

D. Pengelley and R. Laubenbacher
Tue Feb 9 17:35:23 MST 1999