## Sums of Three Cubes

```Fermat's equation for odd exponents n asks for three integers, each
with absolute value greater than 0, such that the sum of their nth
powers is zero.  A related problem is to find three integers, each
with absolute value greater than the nth root of k, such that the
sum of their nth powers equals k.

For example, determine the integers x,y,z  with  |x|,|y|,|z| > 1 such
that
x^3 + y^3 + z^3 = 1                            (1)

This has infinitely many solutions because of the identity

(1 +- 9m^3)^3  +  (9m^4)^3  +  (-9m^4 -+ 3m)^3  =  1           (2)

but there are other solutions as well.  Are there any other identities
that give a different 1-parameter family of solutions?  Is every
solution of (1) a member of a family like this?

In general it's known that there is no finite method for determining
whether a given Diophantine equation has solutions.  However, I wonder
if there is a general method for determining if a given Diophantine
equation has "algebraic" solutions, i.e., an algebraic identity like
the one above that gives an infinite family of solutions.  Does anyone
know of such a method?  (I suppose, in view of Falting's Theorem
(formerly Mordell's Conjecture), that only equations of genus < 2
can have an algebraic solution.)

It may be worth mentioning that the complete *rational* solution of
the equation x^3 + y^3 + z^3 = t^3  is known, and is given by

x = q [ 1 - (a - 3b)(a^2 + 3b^3) ]
y = q [ (a + 3b)(a^2 + 3b^2) - 1 ]
z = q [ (a^2 + 3b^2)^2 - (a + 3b) ]
t = q [ (a^2 + 3b^2)^2 - (a - 3b) ]

where q,a,b are any rational numbers.  So if we set q equal to the
inverse of  [(a^2 + 3b^2)^2 - (a-3b)]  we have rational solutions
of (1).  However, I think the problem of finding the _integer_
solutions is more difficult.  If t is allowed to be any integer
(not just 1) then Ramanujan gave the integer solutions

x = 3n^2 + 5nm - 5m^2
y = 4n^2 - 4nm + 6m^2
z = 5n^2 - 5nm - 3m^2
t = 6n^2 - 4nm + 4m^2

This occassionally gives a solution of equation (1) (with appropriate
changes in sign), as in the following cases

n     m           x       y        z
----  -----     -------  -------  -------
1     -1          (1)       2       -2
1     -2           9       10      -12
5    -12        -135     -138      172
19     -8        -791     -812     1010
46   -109       11161    11468   -14258
73   -173       65601    67402   -83802
419   -993     -951690  -926271  1183258

However, this doesn't cover all of the solutions given by (2).

By the way, my "Most Wanted" problem #16 asks if the equation

x^3 + y^3 + z^3 = 1

has any algebraic solutions other than

(1 +- 9m^3)^3  +  (9m^4)^3  +  (-9m^4 -+ 3m)^3  =  1

and, if so, whether ALL the integer solutions are given by such
an algebraic identity.  Dean Hickerson has informed me via email
that there are known to be infinitely many algebraic solutions,
and he cites the example

(1 - 9 t^3 + 648 t^6 + 3888 t^9)^3 +
(-135 t^4  +  3888 t^10)^3 +
(3 t - 81 t^4 - 1296 t^7 - 3888 t^10)^3  =  1

However, he says it's not known whether EVERY solution of the
equation lies in some family of solutions with an algebraic
parameterization.

He gives the following references

"Sums of three cubes" by G. Payne and L. Vaserstein, in "The
arithmetic of function fields: proceedings of the workshop
at the Ohio State University, June 17-26, 1991", edited by
David Goss, David R. Hayes, and Michael I. Rosen.

"On the Diophantine equation x^3 + y^3 + z^3 = 1" by D. H.
Lehmer, J. London Math. Soc. 31 (1956), 275-280.

Interestingly, Dean notes that if you replace 1 by 2, then again
there's a parametric solution:

(6 t^3 + 1)^3 - (6 t^3 - 1)^3 - (6 t^2)^3  =  2

but again this doesn't cover all known integer solutions, such as

1214928^3 + 3480205^3 - 3528875^3  =  2

It's evidently not known if there are ANY other algebraic solutions
besides the one noted above.  In general it seems to be a difficult
problem to characterize all the solutions of x^3 + y^3 + z^3 = k
for some arbitrary integer k.  In particular, the question of whether
ALL integer solutions are given by an algebraic identity seems both
difficult and interesting.
```