<s>[INST] $\frac{2n+3}{n^2+n+1}$ How many integers are there that make an expression an integer? $$\frac{2n+3}{n^2+n+1}$$
How many integers are there that make an expression an integer?
$-1,0,2$ It is seen that the numbers make the fraction integer immediately.  But the answer key says there are three more $~n~$ values.
 [/INST] If $\dfrac{2n+3}{n^2+n+1}$ is an integer, then it is $0$ or $\ge1$ or $\le-1$.  
But $\dfrac{2n+3}{n^2+n+1}=0$ would mean $n=\dfrac{-3}2,$ not an integer.
And $n^2+n+1>0$ for all $n$, 
so $\dfrac{2n+3}{n^2+n+1}\le-1\implies 2n+3\le-1(n^2+n+1)\implies n^2+3n+4\le0$, 
which is true for no real $n$.
Therefore, $\dfrac{2n+3}{n^2+n+1}\ge1,$ which means $2n+3\ge n^2+n+1$ 
or $n^2-n-2=(n+1)(n-2)\le0$, which means $-1\le n\le 2$.  
Of the integers between $-1$ and $2$, $\dfrac{2n+3}{n^2+n+1}$ is an integer for $n=-1$, $0$, and $2$ but not $1$.
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