How can we find the bounds?

Case 1: $n=k^2$ . $[ \sqrt{n} ] = k$ .

$\dfrac{n^2 + 1}{[\sqrt{n}]^2 + 2}=\dfrac{k^4+1}{k^2+2} = k^2 -2 + \frac{5}{k^2 + 2 }$

For this to be an integer, $\frac{ 5}{ k^2 + 2 }$ must be an integer, which we can check is not possible.

Case 2: $k^2+1 \leq n \leq k^2+k$ then $[\sqrt{n}]=k$ . Let $n = k^2 + r$ with $1 \leq r \leq k$ .

$\dfrac{n^2 + 1}{[\sqrt{n}]^2 + 2}=\dfrac{(k^2+r)^2+1}{k^2+2} = \dfrac{(k^2+2+r-2)^2+1}{k^2+2} =\dfrac{(k^2+2)^2+2(k^2+2)(r-2)+(r-2)^2+1}{k^2+2}$

So, $\dfrac{(r-2)^2+1}{k^2+2}$ should be an integer. However, since $1 \leq (r-2)^2 + 1 < k^2 + 2$ , hence the fraction is strictly between 0 and 1. Thus, there are no solutions.

Case 3: $k^2+k+1 \leq n \leq k^2+2k$ then $[\sqrt{n}]=k+1$ . Let $n = (k+1)^2 - r$ with $1 \leq r \leq k$ .

$\frac{n^2 + 1}{[\sqrt{n}]^2 + 2}=\frac{( (k+1)^2-r)^2+1}{(k+1)^2+2} = \frac{ ((k+1)^2 + 2 - r - 2)^2 + 1 } { (k+1)^2 +2 } = \frac{ ((k+1)^2 + 2)^2 - 2((k+1)^2 + 2 )(r+2) + (r+2)^2 + 1 } { (k+1)^2 + 2 }$

So $\dfrac { (r+2)^2 + 1 } { (k+1)^2 + 2 }$ must be an integer. For $r \leq k-1$ , we have $1 \leq (r+2)^2 + 1 < (k+1)^2 + 2$ , hence the fraction is strictly between 0 and 1.

It remains to check that case of $r = k$ . Substituting this in, we get that $\frac{ k^2 + 4k + 5 } { k^2 + 2k + 3 } = 1 + \frac{ 2(k+1) } { k^2 + 2k + 3 }$ is an integer. However, we see that the fraction is between 0 and 1, hence is never an integer for positive $k$ . Thus, there are no solutions.