From: elkies@brauer.math.harvard.edu (Noam Elkies)
Newsgroups: sci.math.research
Subject: Re: class field theory and diophantine equations
Date: 9 Feb 1997 22:56:47 GMT
Summary: elementary proof of cubic character of 2

In article <5dir4l$gh4@agate.berkeley.edu> vojta@tashkent.berkeley.edu (Paul Vojta) writes:
>I don't know whether you would consider this to be a diophantine equation
>or not, but Ireland and Rosen (as I recall, can't look it up) has the
>following example:

>Let p be a prime number, and let m be the number of distinct
>roots of  X^3-2  [mod p].  Prove the following:

>(a)  m = 0  if and only if  p is congruent to 1 mod 3 but is not of
>the form  a^2+27b^2  with  a,b integers;

>(b)  m=1  if and only if  p is not congruent to 1 mod 3;

[this should actually say "is congruent to 2 mod 3" since m=1 for
p=3 --NDE]

>(c)  m never equals 2; and

>(d)  m=3  if and only if  p=a^2+27b^2 for some  integers a,b.

[I took the liberty of excising some TeX-isms --NDE]

That's a neat example, which nowadays is usually seen as a
manifestation of class field theory.  It should be noted,
however, that the result predates class field theory by
many years (it was conjectured by Euler, and presumably
proved by Eisenstein).  Of course parts (b) and (c)
are elementary and the real question is when does 2
have a cube root mod p when p is congruent to 1 mod 3.

One neat proof of the answer "if and only if p=a^2+27b^2"
uses only the fact that (for p congruent to 1 mod 3) the
number of solutions of the cubic Fermat equation  x^3+y^3=1
mod p  is  p-2+r  where r is the unique integer congruent
to 1 mod 3 such that 4p-r^2 is 27 times a square.  [This
fact, remarkable in itself, can be obtained from the theory
of complex multiplication, from cubic Gauss and Jacobi sums,
or by almost elementary means from a rational parametrization of
the Fermat cubic surface x^3+y^3+z^3=t^3, but that's another post.]

For instance: if p=7 we have r=-1, and if p=13 then r=5, so
in both cases the six solutions with xy=0 are the only ones;
for p=19 we get r=7 so there are 24 solutions, the 6 with
xy=0 and the 18 where {x^3,y^3}={7,-8}; for p=31, r is 4,
so we get 33 solutions: six with xy=0, 18 with {x^3,y^3}={1,-2},
and nine with x^3=y^3=15.

Now the key trick is that there is an involution (x,y)<-->(y,x)
on the set of solutions, whose fixed points are (x,x) with x^3=1/2.
Thus if 2 is not a cube there are no fixed points, and if it is
a cube there are three.  This means that the number of solutions
is odd or even according as 2 is or is not a cube mod 3.  But
that number p-2+r is odd if and only if r is even, i.e. iff
p=a^2+27b^2 for some integers a=r/2 and b, Q.E.D.  For instance
our last case p=31 above is the smallest for which there are
fixed points such as x=y=-8, and also the smallest for which
2 is a cube mod p.

[Exercise: If p is odd, prove by analogous but completely elementary
means that 2 is a square mod p if and only if p is congruent to
1 or 7 mod 8.  If you are more ambitious, prove the law of quadratic
reciprocity in this way: to test whether another odd prime q is a square
mod p, look for fixed points of an order-q permutation of the solutions
of  x1^2+x2^3+...+xq^2=1 mod p.  If you're really ambitious, prove
the full law of cubic and/or quartic reciprocity, starting with the
result that if 3|p-1 then 3 is a cube mod p iff 4p=a^2+243b^2 for
some integers a,b.]

--Noam D. Elkies (elkies@math.harvard:edu)
  Dept. of Mathematics, Harvard University