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


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!)

  • 1
    $\begingroup$ 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. $\endgroup$ – Adam Feb 16 '11 at 22:54
  • $\begingroup$ If $a$ is positive you still get bounded $n,$ after your changes. $\endgroup$ – 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 $a, b$?


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.


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.

  • $\begingroup$ I think you mean c > 3n. $\endgroup$ – Qiaochu Yuan Feb 16 '11 at 17:52
  • $\begingroup$ 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 $\endgroup$ – 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.

Gerhard "Ask Me About System Design" Paseman, 2011.02.16

  • $\begingroup$ In the new version (using s and t), we have s > 0, 2sn +t =a, and ijk -sn^2 = b. With n given, one can get s and t in terms of a and b nicely. (I'll let you do that.) The same general advice holds. Good Luck. Gerhard "Ask Me About System Design" Paseman, 2011.02.16 $\endgroup$ – Gerhard Paseman Feb 16 '11 at 23:13
  • $\begingroup$ Oops: In the comment above, I mean s = (i+j+k) now, not s=(i+j+k+1). Perhaps you should do it yourself. Gerhard "Ask Me About System Design" Paseman, 2011.02.16 $\endgroup$ – Gerhard Paseman Feb 16 '11 at 23:18
  • $\begingroup$ Instead of keeping up with the changes, I'll just suggest the idea above: use n+i for x_1, n+j for x_2, n+k for x_3, or something similar, and see where that takes you. If you develop it and want more suggestions, you can respond here. Again, good luck. Gerhard "Going To Play Elsewhere Now" Paseman, 2016.02.16 $\endgroup$ – Gerhard Paseman Feb 16 '11 at 23:23

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