Diophantine problem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:31:54Z http://mathoverflow.net/feeds/question/55633 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55633/diophantine-problem Diophantine problem Adam 2011-02-16T16:22:17Z 2011-02-16T23:16:24Z <p>I have reduced a knotty research problem to the following reasonable looking form: </p> <p>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 $-3n &lt; x_1+x_2+x_3$, satisfying:</p> <p>$x_1x_2x_3=-n^3-an-b,$ and</p> <p>$x_1x_2+x_1x_3+x_2x_3=a+3n^2.$</p> <p>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?</p> <p>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?</p> <p>(edit: equations corrected. Sorry for time-wasting!)</p> http://mathoverflow.net/questions/55633/diophantine-problem/55638#55638 Answer by Igor Rivin for Diophantine problem Igor Rivin 2011-02-16T16:36:00Z 2011-02-16T16:36:00Z <p>You are asking whether the cubic polynomial</p> <p>$$X^3 - c X^2 + (a + 3 n^2 -2n) X - (n^3 +a n - b) = 0$$ has <em>positive</em> integer solutions under the assumption that $c &lt; 3 n.$ While I don't know the answer, this presumably reduces to standard arithmetic geometry, bypassing Hilbert's tenth problem.</p> http://mathoverflow.net/questions/55633/diophantine-problem/55642#55642 Answer by Charles Matthews for Diophantine problem Charles Matthews 2011-02-16T16:53:28Z 2011-02-16T16:53:28Z <p>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 <em>n</em>. I think this comes out bounding <em>n</em> in terms of <em>a</em> and <em>b</em>. And unless the x-values are similar in size, there should be more. But most <em>n</em> don't factor like that, so I would expect this to fail.</p> http://mathoverflow.net/questions/55633/diophantine-problem/55645#55645 Answer by Qiaochu Yuan for Diophantine problem Qiaochu Yuan 2011-02-16T17:10:48Z 2011-02-16T17:47:00Z <p>Following up Charles Matthews' idea, <a href="http://en.wikipedia.org/wiki/Maclaurin%27s_inequality" rel="nofollow">Maclaurin's inequality</a> gives</p> <p>$$\frac{x_1 + x_2 + x_3}{3} \ge \sqrt{ \frac{3n^2 - 2n + a}{3} } \ge \sqrt[3]{ n^3 + an - b}.$$</p> <p>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. </p> <hr> <p>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 $a, b$?</p> http://mathoverflow.net/questions/55633/diophantine-problem/55665#55665 Answer by Gerhard Paseman for Diophantine problem Gerhard Paseman 2011-02-16T21:14:42Z 2011-02-16T21:24:06Z <p>Here is an alternative formulation (possibly your original one) where $x_m$ is replaced by $n +$ something which yields $0 &lt; i+j+k$ with each of $i,j,k \ge -n$ . Then (I've already fixed one mistake, so check my work)</p> <p>$2(i+j+k+1)n + (ij+jk +ki) = a$</p> <p>$(i+j+k)n^2 + (ij+jk+ki)n +ijk = an - b$</p> <p>$(i+j+k+2)n^2 - ijk = b$</p> <p>Since $(ij +jk +ki)$ can be negative, we don't have $a > n$ or even $b> 0$.</p> <p>However there are inequalities mentioned in other posts which apply to the terms $(s-1) = (i+j+k)$ and $t =(ij +jk +ki)$. <strike>Further, one has $an/2 - b = ijk$.</strike> 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.</p> <p>Gerhard "Ask Me About System Design" Paseman, 2011.02.16</p>