Like Diophantine equation

2013-02-12T05:01:14Z

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.

The equation $x^n - ny^x-nxy$ = $0$ has solution set $(n, x, y) = (1, 1, \frac12), (2, 1, \frac14), (3, 1, \frac16), \ldots$

I would like to know/learn the following (Kindly discuss)

1) If we want to know the graph. How would be the look of the graph and what kind of graph we get?

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?

3) If solutions exists how to find them for $x < 1$ and $x > 1$?

Thanks in advance.

Answer by Max Alekseyev for Like Diophantine equation

2013-03-03T08:28:07Z

The equation $x^n - ny^x - nxy = 0$ has no solutions in positive integers.

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.

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 $u>0$.

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 $u+v+w < xv+w$ unless $v=0$. So there are two cases to consider:

1) $v=0$. In this case we have $xv+w = w < u+v+w$ and $xv+w = w < p^w \leq nu$, that is $\nu_p(ny^x)$ is a sole smallest valuation among the three, which is impossible.

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.

Like Diophantine equation - MathOverflow

Like Diophantine equation

Answer by Max Alekseyev for Like Diophantine equation