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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.

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@m.s.e. memebrs. Kindly answer my post. – jihadi Feb 12 '13 at 3:59
up vote 1 down vote accepted

Let's take $x=2$ and see what happens.

The equation becomes $ny^2+2ny-2^n=0$. That's a quadratic equation for $y$, so we get $$y={-2n\pm\sqrt{4n^2+(4n)2^n}\over2n}={-n\pm\sqrt{n^2+2^nn}\over n}$$ Wanting $n$ and $y$ to be positive, we reject the negative root, and get $$y={\sqrt{n^2+2^nn}-n\over n}$$ so there is a $y$ for every $n$, hence, an uncountable infinity of solutions for $x=2$.

$x=1/2$ also leads to a quadratic in $y$ --- details left to the reader.

If by "the graph" we mean the set of all points $(n,x,y)$ in ${\bf R}^3$ satisfying the equation, I doubt it looks like much of anything.

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! you have made an excellent note for x = 2 as well as x = 1/2. I am agree with your argument on graph. But, can you show me the graph on $R^3$. Also, for x = 2 case, we have seen for any n, we get some y. is there any restriction on n, to get a positive y? as per your terms, for any n, we get y. Will we get for any positive n, the value of y is positive? also, integer? – jihadi Feb 12 '13 at 5:59
Gerry Myserson! once again, I am so thankful to your for your valuable observation on x = 2 and x = 1/2. – jihadi Feb 12 '13 at 6:01
Gerry Myerson! for n =1, we get y in rationals. But, we will have some solutions. No doubt! on it. can we find solutions in integers? at least one solution?I think, we cant. But, I need your advise. Since, your analysis is exceptional – jihadi Feb 12 '13 at 6:02
For $x=2$, you get a positive $y$ for every positive $n$ --- just look at the formula! For $x=2$, I you can certainly get $y$ to be an integer, but I don't think you can get $n$ and $y$ both integers. I don't know how to show you the graph --- I recommend you find some graphing software, and let it do the work. – Gerry Myerson Feb 12 '13 at 6:23
Gerry Myerson! Thank you for your quick reply on my post. You are absolutely correct, we will have solutions for any n some y. For any n, can we get some x (other than 1 and 2)? – jihadi Feb 12 '13 at 6:30

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