How to create solvable equations with algebraic fractions? I came across this simple algebraic fractions question:
$\frac{8}{x+3}+\frac{3}{x+8}=1$
After a bit of manipulation you are left with a difference of two squares and it is easily solvable.
I then wondered how these questions are made, so had a go at creating a new one and it has been surprisingly difficult.
My first thought was to work backwards from a factorisable quadratic, but I hit a dead end.
My attempt now is to begin with something in the form:
$\frac{a}{x+b}+\frac{b}{x+a}=1$
Which leads to:
$0=x^2+ab-a^2-b^2$
So my question is, how to pick values of a and b so that this will be factorisable, or more specifically so that it will be a difference of two squares? i.e:
$ab-a^2-b^2=-n^2$ for some integer n.
 A: To elaborate on lulu's perfect answer:
$a^2+b^2-ab=n^2\iff x^2+y^2-xy=1$, where $x=a/n, y=b/n$.
Now, to parametrize the points of this ellipse, you can start from the point $I=(1,0)$, and then intersect a line of $\mathbb{R}^2$, passing through $I$. The equation of such a line has the form $y=t(x-1)$. Plugging into the ellipse equation yields $x^2+t^2(x-1)^2- tx(x-1)-1=0$, that is $(x-1)(x+1+t^2(x-1)-tx)=0$.
The case $x=1$ will give the point $I$. If $x\neq 1$, little algebraic computations yield $x=\dfrac{t^2-1}{t^2-t+1}, y=\dfrac{t^2-2t}{t^2-t+1}$.
If you take for $t$ any rational value, you will get rational points on your ellipse (in fact, one may convince ourselves that we get all of them).
Side remark. Letting $t$ go to $+\infty$ yield the missing point $I$.
Multiplying $x,y$ by a suitable integer $n$ (for example a denominator of $x,y$), you will get integers $a,b$ as you want.
An easy infinite family of examples would be $a=m^2-1, b=m^2-2m$, where $m$ is an arbitrary integer.
A little algebra shows that  $(m^2-1)^2+(m^2-2m)^2-(m^2-1)(m^2-2m)=(m^2-m+1)^2$, so for $a=m^2-1, b=m^2-2m$, you may take $n=m^2-m+1$.
Note that the case $a=8$ and $b=3$ corresponds to $m=3$.
