what points make $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ true? I was asked the following question: 
for each integer $n>1$, find distinct positive integers $x$ and $y$ such that 
$\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$
This is what I got so far: 
Doing some algebra we can see that $(x,y)$ have to be integers such that $\frac{xy}{x+y}=n$ and from here we can obtain the following expression $y=n(\frac{x}{x-n}) $.From this expression we can see some stuff like $x>n$. By playing with it for a bit, I reached the conclusion that for each $n$ the $(x,y)$ that work are (n+1,n(n+1)),(2n,2n). by plugging them in into $y=n(\frac{x}{x-n}) $  we can see that these points work however I do not know how to show these points are the only points that work. 
Thank you, any help is appreciated. 
 A: Since $x,y,n$ are positive integers we must have $x>n$ and $y>n$. Let $x=n+s$ and $y=n+t$ for some appropriate $s,t\in\mathbb{N}$. Then your equation becomes $\frac{1}{n+s}+\frac{1}{n+t}=\frac{1}{n}$ or by cross multiplications we finally get $n^2=st$. Hence we arrive at an easier equation to solve.
In fact, you can check there are exactly $\tau(n^2)$ solutions $(s,t)$ to the equation $n^2=st$.
Happy Problem-solving
Enjoy problem-solving and fight Corona
A: They aren't the only ones that do work.
Well, here's my solution:
$$\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$$
$$\iff x+y=\frac{1}{n}\cdot xy$$
$$\iff x(1-\frac{y}{n})+y=0$$
$$\iff \frac{x}{n}(n-y)+y=0$$
$$\iff -\frac{x}{n}(y-n)+y-n=-n$$
$$\iff (1-\frac{x}{n})(y-n)=-n$$
$$\iff (x-n)(y-n)=n^2$$
So, let $a$ be an arbitrary positive divisor of $n^2$, then
$$x \in S = \{a+n: a\ |\ n^2\}$$
$$y \in \{\frac{n^2}{x-n}+n:x\in S\}$$
Or 
$$x \in S = \{-a+n: a\ |\ n^2, a \ne n\}$$
$$y \in \{\frac{n^2}{n-x}+n:x\in S\}$$
For example, some known solutions are $(x,y) \in \{(n^2+n,n+1),(n+1,n^2+n),(n-n^2,n-1),(n-1,n-n^2),(2n,2n)\}$
A: All solutions of $yn+nx-xy=0$  with $\gcd(x,y,n) = 1$ and $xyn >0$ come from $\gcd(u,v) = 1$  and $$ x = u^2 + uv \; , \; \; y = uv + v^2 \; , \; \; n = uv $$   as the quadratic form is "equivalent" to the Fricke-Klein form $s^2 - tr.$
It takes just a few lines and some gcd calculations to show that if $a > 0$   and $c^2 = ab$   and $\gcd(a,b,c) = 1,$  then there are $u,v$ with $\gcd(u,v) = 1$ such that $a = u^2, c = uv, b = v^2.$   We may also negate all three.
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Take integers $a,b,c$   with $a > 0,$  then $\gcd(a,b,c) = 1,$  finally $ab=c^2$
Take $$ g = \gcd(a,c) $$ so that
$$ a = g \alpha \; , \; \;  c = g \gamma   $$
with $\gcd(\alpha, \gamma)=1.$  Then $ab=c^2$  gives $g \alpha b = g^2 \gamma^2$ or
$$   \alpha b = g \gamma^2 $$
Since $\alpha, \gamma $ are coprime, we get two division facts, $\alpha | g$  and $\gamma | b.$ From the latter we name $b = h \gamma,$   and $ab=c^2$  gives us
$$  \alpha h = g \gamma . $$
From the forme, name $g = j \alpha.$  Right away we get $h = j \gamma.$
We now have
$$  a = j \alpha^2 \; , \; \;  b = j \gamma^2 \; , \; \; c = j \alpha \gamma  $$
As $\gcd(a,b,c) = 1,$  we know $j = \pm 1.$  Finally, we demanded $a > 0,$   so that $j=1.$ We have provided coprime integers with
$$  a =  \alpha^2 \; , \; \;  b =  \gamma^2 \; , \; \; c =  \alpha \gamma  $$
