On the equation a^m+b^n=c^p

Let me from memory post all that one knows about the general
equation a^m+b^n=c^p in integers. One of the main world experts
on the subject is F. Beukers, but I would be surprised if he read sci.math!

1) If a, b, c are NOT coprime, you can trivially construct an infinite
number of solutions, even when m, n, p are distinct. For example,
assume (almost random example) that you want a solution with m=3, n=5,
p=7. Choose n1=84, n2=90 (see why?). Then, since 1+2=3, you obtain
a=2^28.3^30, b=2^17.3^18, c=2^12.3^13 as a solution.

So assume from now on that a, b, c are coprime (by pairs or globally,
this is equivalent). Assume in addition a, b and c nonzero.
We distinguish 3 cases.

2) The hyperbolic case: this is when 1/m+1/n+1/p<1. By a simple counting
argument, one EXPECTS that there is only a finite number of solutions.
That this is the case is a very deep theorem only proved 2 years ago by
Darmon and Granville: they are able to reduce the problem to Faltings's
result on the Mordell conjecture, but the reduction is highly non trivial
because the equation is not homogeneous.

Computer calculations over a very large range have given exactly 13
(I think) solutions (counting as a single one 1^m+2^3=3^2), some of them
quite large, and all of them with one of the exponents equal to 2.
It is conjectured that there are only finitely many in total (Darmon-
Granville only prove that it is finite for FIXED m,n,p). In fact very possibly
there are no more than those discovered, or perhaps 1 or 2 more.
I have the list at home for those interested (as well as all the references
I do not give), but go back home only in 3 weeks!

3) The parabolic case: this is when 1/m+1/n+1/p=1. This is the easiest, and
in a sense has been completely treated by Fermat. It includes for example
a^3+b^3=c^3, a^4+b^4=c^2, etc... Here the answer is simple: there are no
solutions (recall we assume a,b,c non zero).

4) The elliptic case: this is when 1/m+1/n+1/p>1, and includes the equation
a^4+b^3=c^2 considered in this thread. A counting argument shows that
it SHOULD have an infinite number of solutions. This is indeed the case,
and has been proved by F. Beukers. However his result is much more
spectacular, and answers positively the question asked by P. Lounesto:
For given elliptic (m,n,p), the set of solutions is COMPLETELY described
by a FINITE family of polynomial parametric solutions, where the polynomials
are homogeneous (of known degree) with 2 variables. These families do NOT
intersect, except trivially, so the complete solution set is known.
The question of finding explicitly these polynomials is another matter.
However, the elliptic case boils down to only a small number of cases.
Let (u,v,w) be (m,n,p) ordered so that u<=v<=w. We can exclude the
trivial case where u=1, so assume u>=2. Then we have the following cases:
u=2, v=2, w>=2 : this can be called the dihedral case of order w.
u=2, v=3, w=3: the tetrahedral case
u=2, v=3, w=4: the octahedral (or cubic) case 
u=2, v=3, w=5: the icosahedral (or dodecahedral) case.
The relation with platonic solids is not a coincidence as I will explain.

Anyway, to summarize, Zagier and Beukers have found the complete parametric
set in all the elliptic cases, except for the icosahedral case. In the latter,
already 22 independent parametric families are known, but there may be many more.
In particular, the complete parametric solution of the equation of the thread
(a^4+b^3=c^2) is known, but the explicit form is at home (you know, the margin
of this post is too small....)

Now let me explain what all this has to do with the platonic solids by taking
the example of the thread a^4+b^3=c^2.

We are going to do a little magic, but believe me when I tell you the magic
works also in the other platonic cases.

a) Draw in space a regular ocathedron symmetric with respect to the z=0 plane,
whose interection with the z=0 plane is the square formed by the 4 fourth
roots of unity (+1, -1, +i, -i). This is not essential, but just to fix the
picture, which evidently I am unable to draw in ASCII.
Take one of the vertices not on the z=0 plane as pole and project all the
vertices from there (the standard steregraphic projection). Let z_k be the
6 points (in the projective plane, but if you don't know what this is don't
worry) thus obtained: they simply are +1,-1,+i,-i,0,infinity.
Now call f the two variable homogeneous polynomial having the z_k as simple
roots: here this is simply x*y*(x^4-y^4) (infinity corresponds to the root y=0).
So f(x,y)=x*y*(x^4-y^4).

Now more magic: let h(x,y) be the Hessian of f, i.e. fxx*fyy-fxy^2 where
fx, fy, etc... denotes derivation with respect to x, y, etc...
Here we see that it is divisible by (-25) so we set in fact h to be equal
to (-1/25) times the Hessian, explicitly h(x,y)=x^8+14*x^4*y^4+y^8.

Let j(x,y) be the Jacobian of f and h, i.e. fx*hy-fy*hx. This is divisible
by (-8) so we set in fact j equal to (-1/8) times the Jacobian, explicitly
j(x,y)=x^12-33*y^4*x^8-33*y^8*x^4+y^12.

The punchline and final piece of magic: -108*f^4+h^3=j^2 !!!

Now this isn't our equation a^4+b^3=c^2, but it is pretty close, and
it is easy to get rid of the annoying -108: for instance, set formally
x=s*4^(1/4), y=t*(-3)^(1/4). Then, since h and j involve only x^4 and y^4,
only integral coefficients will change, while in f^4 an extra factor
of -12 will come out, and (-12)*(-108)=6^4. Thus, we have the explicit
parametrization for a^4+b^3=c^2:

a=6*s*t*(4*s^4+3*t^4)
b=16*s^8-168*s^4*t^4+9*t^8
c=64*s^12+1584*s^8*t^4-1188*s^4*t^8-27*t^12

I seem to remember that there is at least another one.

Easy exercise: do the same for the tetrahedron.
Slightly more difficult exercise: for the icosahedron.

	Enjoy!

		Henri Cohen