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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.

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    $\begingroup$ Why not cancelling $x-2$ ?? $\endgroup$ – IrbidMath Oct 24 '19 at 1:10
  • $\begingroup$ If you cancel then $\frac{x-2}{x+3}=1+\frac{-5}{x+3}$ $\endgroup$ – IrbidMath Oct 24 '19 at 1:25
  • $\begingroup$ @AmerYR I understand. Or you could say $\frac{x-2}{x+3}=k$ for some integer $k$ and then solve and get $x=\frac{2+3k}{1-k}$. So, again, we will get a formula for $x$, but actually this formula is better than what I got in my question. So maybe this is the solution itself. $\endgroup$ – sequence Oct 24 '19 at 1:26
  • $\begingroup$ We get just $x=-8$ the other choice is impossible zero of the denominator $\endgroup$ – IrbidMath Oct 24 '19 at 1:28
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    $\begingroup$ @AmerYR While your observation does help for this problem, I'm wondering if there is a more general approach to this sort of problems. Suppose we couldn't factor and cancel any factors, then how would we proceed? $\endgroup$ – sequence Oct 24 '19 at 1:30
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$\frac{x^2-4x+4}{x^2+x-6}=\frac{x-2}{x+3}=1-\frac{5}{x+3}$, 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.

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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.

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    $\begingroup$ This gives integer solutions, but not all $x$, because there are also rational solutions (for example, $x=-\frac12, x=-\frac{11}{2}$, etc). $\endgroup$ – sequence Oct 24 '19 at 23:25
  • $\begingroup$ You're right, I've overlooked the "rational". $\endgroup$ – user376343 Oct 25 '19 at 7:55
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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 .

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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.

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  • $\begingroup$ In the comments I also wrote that the solution can be expressed as $x=\frac{2+3k}{1-k}$, which seems to be more straightforward. $\endgroup$ – sequence Oct 24 '19 at 2:01
  • $\begingroup$ Yes, that seems even better if that expression covers all the cases covered by expression in this answer. $\endgroup$ – user716491 Oct 24 '19 at 2:03

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