Diophantine problem I have reduced a knotty research problem to the following reasonable looking form: 
Given any two integers $a$ and $b$, show that there are natural numbers $x_1,x_2,x_3$ and a (probaby negative) integer $n$, where $-3n < x_1+x_2+x_3$, satisfying:
$x_1x_2x_3=-n^3-an-b,$ and
$x_1x_2+x_1x_3+x_2x_3=a+3n^2.$
I am not expecting a solution to this (although that would of course be the ideal outcome)!  However, I don't really know where to start.  How might one go about solving something like this?  Are there any tried and tested methods I should know about?
And finally, given the unsolvability of Hilbert's tenth problem, is it possible that there is no way to know whether or not this is true?
(edit: equations corrected.  Sorry for time-wasting!)
 A: My guess is that it doesn't work. But I think elementary methods are your friend here. For example the two equations seem set up to apply the AM-GM inequality here, which apparently yields a comparison of two sextic polynomials in n. I think this comes out bounding n in terms of a and b. And unless the x-values are similar in size, there should be more. But most n don't factor like that, so I would expect this to fail.
A: Following up Charles Matthews' idea, Maclaurin's inequality gives
$$\frac{x_1 + x_2 + x_3}{3} \ge \sqrt{ \frac{3n^2 - 2n + a}{3} } \ge \sqrt[3]{ n^3 + an - b}.$$
The second inequality in particular expands out to an inequality of the form $-54n^5 + \text{lower order terms} \ge 0$, so does in fact provide an upper bound for $n$ in terms of $a$ and $b$. If you don't expect the statement to be true, from here it is possible to search for counterexamples. 

If I'm not mistaken, the above inequality never holds when $a = b = 1$, so no such $n$ exists in this case. In general in order to get a reasonable number of possibilities for $n$, $a$ needs to be large compared to $b$. Are you sure you meant to ask the question about any possible $a, b$?
A: You are asking whether the cubic polynomial
$$ X^3 - c X^2 + (a + 3 n^2 -2n) X - (n^3 +a n - b) = 0$$ has positive integer solutions under the assumption that $c < 3 n.$ While I don't know the answer, this presumably reduces to standard arithmetic geometry, bypassing Hilbert's tenth problem.
A: Here is an alternative formulation (possibly your original one) where
$x_m$ is replaced by $n +$ something which yields $0 < i+j+k$ with each of
$i,j,k \ge -n$ . Then (I've already fixed one mistake, so check my work)
$2(i+j+k+1)n + (ij+jk +ki) =  a$
$(i+j+k)n^2 + (ij+jk+ki)n +ijk = an - b$
$(i+j+k+2)n^2  - ijk = b$
Since $(ij +jk +ki)$ can be negative, we don't have $a > n$ or even $b> 0$.
However there are inequalities mentioned in other posts which apply to
the terms $(s-1) = (i+j+k)$ and $t =(ij +jk +ki)$.  Further, one has
$an/2  - b = ijk $.  So it might be useful to rewrite the system using
$s$ and $t$ and solve it
 given $n$, and then see if $i,j,k$ can be found after that.
Gerhard "Ask Me About System Design" Paseman, 2011.02.16
