Show that for any positive integer $n>1$ there exist positive integers $x$ and $y$ such that $n^3+x^3=(x-1)^3+y^3$.

Question

A quick check shows that $x=y=n+1$ work pretty good. Since,

$n^3+(n+1)^3=\{(n+1)-1\}^3+(n+1)^3.$

How to check whether there exist any other type of solutions or not? Please suggest.. Thanks in advance.

Answer 1

There are other solutions such as $x=n(3n^2-3n+2), y=3n^2-2n+1$. To find these, and others, notice that the original equation can be written \begin{equation*} 3x^2-3x+n^3+1=y^3 \end{equation*} which is just an (unusual) elliptic curve. Straightforward algebra gives the more usual form \begin{equation*} v^2=u^3-432(4n^3+1) \end{equation*} where $u=12y$ and $v=72x-36$. Thus, the solution quoted, $x=y=n+1$, corresponds to the point $u=12(n+1), v=36(2n+1)$.

Investigations of these elliptic curves for small $n$ shows no finite torsion points, but the rank is at least $2$. Further investigations give the point $(12(3n^2-2n+1),36(6n^3-6n^2+4n-1))$ as a point, which gives the solution quoted at the beginning.

Other points can come from further investigations.

Answer 2

$a^3+b^3=c^3+d^3$

$a=3(3p^2-17p+5)$

$b=3(85p^2-63p+11)$

$c=2(128p^2-94p+17)$

$d=2(-29p^2+4p+1)$

For $p=(1,-1,0)$ we have

$(a,b,c,d)=(9,34,33,16)=(-64,478,477,75)=(2,34,33,15) $

Answer 3

From a known numerical solution more solutions can be arrived at.

Let $(a,b,c,d)$ be a known solution. We parameterize:

$a^3+b^3=c^3+d^3$ at,

$[(a+k),(b+t),(c+t),(d+k)]$

And we get for $(a,b,c,d)=(2,16,15,9)$

$t=w(31w+77)/(7-w^2)$

$k=w (11w+31)/(7-w^2)$

For $w=(1,3,-2,-3)$ we get:

$(p,q,r,s)=(9,34,33,16)=(94,164,165,87)=(-4,6,5,3)=(-6,9,8,1)$