Like Diophantine equation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T00:50:24Z http://mathoverflow.net/feeds/question/121557 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/121557/like-diophantine-equation Like Diophantine equation jihadi 2013-02-12T05:01:14Z 2013-04-14T10:22:00Z <p>Dear all, I have posted this question on m.s.e. Unfortunately, no one responded to answer. I hope, this site and members of this site will answer my questions. </p> <p>The equation $x^n - ny^x-nxy$ = $0$ has solution set $(n, x, y) = (1, 1, \frac12), (2, 1, \frac14), (3, 1, \frac16), \ldots$</p> <p>I would like to know/learn the following (Kindly discuss)</p> <p>1) If we want to know the graph. How would be the look of the graph and what kind of graph we get?</p> <p>2) The cited above equations has infinite solutions with $x = 1$. Can we have solutions with $x >1$ and other $n, y$ are some positives?</p> <p>3) If solutions exists how to find them for $x &lt; 1$ and $x > 1$? </p> <p>Thanks in advance.</p> http://mathoverflow.net/questions/121557/like-diophantine-equation/123465#123465 Answer by Max Alekseyev for Like Diophantine equation Max Alekseyev 2013-03-03T08:28:07Z 2013-03-03T08:28:07Z <p>The equation $x^n - ny^x - nxy = 0$ has no solutions in positive integers. </p> <p>First notice that $ny$ must divide $x^n$, and $x$ in turn must divide $ny^x$. Therefore, the set of prime divisors of $x$ and $ny$ is the same.</p> <p>Let $p$ be any prime dividing $x$ (or $ny$) and $u=\nu_p(x)$, $v=\nu_p(y)$, $w=\nu_p(n)$ be the corresponding $p$-adic valuations. We remark that $u>0$.</p> <p>Since $x^n - ny^x - nxy = 0$, the two smallest values among $\nu_p(x^n)=nu$, $\nu_p(ny^x)=xv+w$, $\nu_p(nxy)=u+v+w$ must be equal. It is easy to see that $u+v+w &lt; xv+w$ unless $v=0$. So there are two cases to consider:</p> <p>1) $v=0$. In this case we have $xv+w = w &lt; u+v+w$ and $xv+w = w &lt; p^w \leq nu$, that is $\nu_p(ny^x)$ is a sole smallest valuation among the three, which is impossible.</p> <p>2) $v>0$ and $u+v+w = nu$, that is, $v+w=(n-1)u$ and thus $\nu_p(ny) = \nu_p(x^{n-1})$. Since $p$ is arbitrary prime dividing $x$ and $ny$, we conclude that $ny = x^{n-1}$. The equation take form: $$x^n - x^{n-1}y^{x-1} - x^n = 0$$ which reduces to $$x^{n-1}y^{x-1} = 0,$$ a contradiction.</p>