三个立方数的和

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.

引用

\begin{Example}
(\href{https://artofproblemsolving.com/community/c103441_2005_austrianpolish_competition}{2005 Austrian-Polish Competition 3})考虑方程$x^3 + y^3 + z^3 = 2$.
\begin{enumerate}
\item 证明该方程有无穷多个整数解$x,y,z$;

\item 确定所有的整数解$x, y, z$,满足$|x|, |y|, |z| \leq 28$.
\end{enumerate}
\end{Example}
$$\begin{Proof} 注意到$(x,y,z)=(1+6t^3+1,1-6t^3,-6t^2)$是该方程的解,且$2 = 1214928^3 + 3480205^3 - 3528875^3$是第一个满足上述表达式的解. \end{Proof}$$

$$\begin{Example} 求$(a_0,a_1,a_2)\in\mathbb{R}^3$,使得$\int _ { 0 } ^ { 1 } \left| e ^ { 2 } - a _ { 0 } - a _ { 1 } t - a _ { 2 } t ^ { 2 } \right| ^ { 2 } d t$取最小值. \end{Example}$$
\begin{Proof}

\end{Proof}

$$\begin{Example} 设$f(x)\in C^2[a,b]$,满足边界条件: \[f ( a ) = f ( b ) = 0 , \quad f' ( a ) = 1 , \quad f' ( b ) = 0.\] 求证: \[\int _ { 0 } ^ { b } \left| f ^ { \prime \prime } ( x ) \right| ^ { 2 } d x \geq \frac { 4 } { b - a }.\] \end{Example}$$
\begin{Proof}

\end{Proof}

$$\begin{Example} 求证:\[\int_{0}^{1}\frac{\ln (1+x)}{1+x^2}dx=\frac{\pi\ln2}{8}.\] \end{Example}$$
\begin{Proof}