2-variable Quartic Diophantine equation about number sequences How could I prove that $$7n^4 + (12s + 6)n^3 + (6s^2 + 6s + 1)n^2 - 2s^3 = 0$$ has no natural solutions?
The solutions lead to summation identities and solving this one would prove there are no cubic sequences. On the other hand, any proof that there are no cubic sequences would also prove this has no solutions.
I felt that since $s^3$ is the only negative term and it is dominated by $n^4$ there may be hope in solving this.
 A: Define the two variable polynomial
$$p(n,s)=7n^4+(12s+6)n^3+(6s^2+6s+1)n^2-2s^3.$$
Now make the change of variables $n=r,s=m-r$, and note this is an invertible change since we have $r=n,m=n+s$ as
the way to get back from $(r,m)$ to the original variables $(n,s)$. Now $p$ has a simpler formula in the variables $r,m$, 
namely we have
$$p(r,m-r)=r(r+1)[ r(r+1)+6m^2 ] - 2m^3.$$
Now if this were zero for some integers $r,m$ we would have
$$2m^3 = r(r+1)[r(r+1)+6m^2],$$
and then rearranging and factoring we would have
$$ 2m^2 [m-3r(r+1)]=[r(r+1)]^2.$$
Then since the right side here is positive we would have $m>3r(r+1)$, from which 
$$2m^2 >18 [r(r+1)]^2,$$
which is not possible because $2m^2$ is actually a factor of $[r(r+1)]^2$ in the previous equation.
Note: The change of variables here really amounts to using $m$ as the final cube in the longer sequence of cubes to
the left of the proposed equation in the original search for examples like $3^3+4^3+5^3=6^3+7^3$ (which is false), so that 
the search is for consecutive sequences of cubes having total length odd, where the middle cube
plus the ones before it sum to the same as the higher cubes.
A: Taking your item above as $f(n,s),$ with $n \geq 0$ I get
$$  f(-1 - n, s + 2 n + 1 ) = f(n,s).    $$
Just a thought.
