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!)



Diophantine problem
Following up Charles Matthews' idea, [Maclaurin's inequality](http://en.wikipedia.org/wiki/Maclaurin%27s_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. 




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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 $a, b$?


