# 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!)

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Thanks! To everyone. Your observations have made me realise I've made a stupid mistake. I have corrected the equations in my question...they now look a lot more probable. –  Adam Feb 16 '11 at 22:54
If $a$ is positive you still get bounded $n,$ after your changes. –  Will Jagy Feb 16 '11 at 23:06

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 $a, b$?

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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.

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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.

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I think you mean c > 3n. –  Qiaochu Yuan Feb 16 '11 at 17:52
with the correction it should be possible with a=2 and b=1 to find the local maximum of the cubic and verify that it is below the x axis –  Aaron Meyerowitz Feb 16 '11 at 23:17

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.