# Like Diophantine equation

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 $x > 1$?

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A Diophantine equation is typically one where one is interested in integeral solutions only. Now you mention solutions containing rational numbers. It is now not clear what exactly you are looking for. Please clarify. –  quid Feb 12 '13 at 12:34
It is not Diophantine equation. However, I felt it is like Diophantine. Can we have solutions in integers or in other than integers? how to solve such equations? may I know the look of the graph of this function? –  jihadi Feb 12 '13 at 16:36
If you want to know how the graph looks like (of what function even?, but I guess the one devined by the expression on the left of the equation), why not just plot it (using [freely] available software) for selected values of n fixed as a function of x and y? An what do you mean by 'other than integers' precisly? Say fix x,n as some integers. So you have a polynomial in y, which of course has solutions (in the complex numbers at least). Yet I somehow doubt this is what you want to know. So what is it you want to know? –  quid Feb 12 '13 at 20:17
I am unable to draw the function by graph. Moreover, I am looking the solutions in integers. If there, how to find such integer solutions? Is there any particular method to obtain integers solutions? Other wise solutions in $R^3$. –  jihadi Feb 13 '13 at 4:08

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.

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