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