Find $x$ so that rational function is an integer Find all rational values of $x$ such that $$\frac{x^2-4x+4}{x^2+x-6}$$ is an integer.
How I attempt to solve this: rewrite as $x^2-4x+4=q(x)(x^2+x-6)+r(x)$. If we require that $r(x)$ be an integer then we can get some values of $r(x)$ by solving $x^2+x-6=0$, so that $x=-3$ or $x=2$. In the former case, $r=25$, and in the latter case $r=0$.
[I should note, however, that I'm not really convinced that this step is actually correct, because when $x$ is a root of $x^2+x-6$, the rational function above can have no remainder due to division by $0$. But I read about it as a possible step here and I can't explain it. Q1 How can this be justified?]
Now we want $(x^2+x-6)$ to divide $0$ (trivial) or $x^2+x-6$ to divide $25$. So we set $x^2+x-6=25k$ for some integer $k$ and solve. So $$x=\frac12 (\pm5\sqrt{4k+1}-1)$$
One of the trial-and-error substitutions for $k$ and then for $x$ gives $x=-8$, but that is just one number. Q2: So I'm wondering, how can we find all such $x$? The condition here is that $4k+1$ must be a perfect square, $k\ge 0$. Aren't there infinitely many perfect squares of this form?
I'd appreciate some clarifications about these two questions, Q1 and Q2.
 A: $\frac{x^2-4x+4}{x^2+x-6}=\frac{x-2}{x+3}=1-\frac{5}{x+3}$, for this to be an integer, $\frac{5}{x+3}$ has to be an integer.
Say, $\frac{5}{x+3}=K$, where $K \in \Bbb Z$. This implies for each $K \in \Bbb Z$,  $x=\frac{5}{K}-3$ would make the given expression an integer.
A: For $x \neq 2$ is $$\frac{x^2-4x+4}{x^2+x-6}=\frac{x-2}{x+3}=1-\frac{5}{x+3}.$$
To obtain an integer with an integer $x,\;$ $x+3$ must be a divisor of $5,$ or equivalently $x+3 \in \{-1,1,-5,5\}.$
This gives the solutions $-4,-2$ and $8,$ because $x=2$ is excluded.
A: If we want $x$ rational such that  $ \frac{x^2-x+3}{x^2+5x-7}$ is integer I will simplify $1+\frac{-6x+10}{x^2+5x-7}$
Then I will 
go let $a/b$ with  gcd 1 then I will search for solutions to $a^2+5ab-7b^2\mid -6ab+10b^2$ 
Number theory . 
A: The set of all solutions can be described as $5n=k(m+3n)$, where $m \neq -3n$ and $m \neq 2n$ and $m,n,k \in \mathbb Z$. When solutions of the above are plugged into $x= \dfrac {m}{n}$ then those x´s are the required ones.
