With 2 as smallest period of the function $f(x)$= $\tan^2[(\frac{\pi x}{n^2-5n+8})]$ + $\cot(n+m)\pi x$ ;the period m can't belong to is?

Question

Here n $ \in N$ , m $\in Q$.
Options are:
A) $(-\infty, -2) \cup (-1, \infty)$
B) $(-\infty, -3) \cup (-2, \infty)$
C) $(-2,-1) \cup (-3,-2)$
D) $(-3, -5/2) \cup (-5/2, -2)$

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I have an answer to this question in my book but I don't understand it.
It is:

Period is LCM of $n^2-5n+8$ and $\frac{1}{n+m}$ .
Since $n\in N$
solving $n^2 -5n +8=2$ and $n^2 -5n +8=1$ gives $n \in ${2,3}

and $(\frac{1}{n+m})(K) = 2$ ,where K is an integer
$\Rightarrow$ $(\frac{1}{n+m})\ngtr 1 $
$\Rightarrow$ $(\frac{1}{2+m})\ngtr 1 $ or $(\frac{1}{3+m})\ngtr 1 $
$\Rightarrow$ $m \notin (-2,-1) \cup (-3,-2)$

Would anyone please explain how did:
$(\frac{1}{n+m})(K) = 2$$\Rightarrow$ $(\frac{1}{n+m})\ngtr 1 $ happen?

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If there is any other way then please post it. Thankyou.