Prove that there is always an even number in the interval $[ \sqrt{9 + 8n} - 3, \sqrt{1 + 8n} - 1]$ for all positive integers n I found this problem on a PDF that I could not find the solutions for.
I initially tried thinking about the size of the interval, however it is always smaller than 2 and therefore does not guarantee an even number.
I thought that maybe the left hand boundary will always be a little bit above an odd number and as the interval is eventually larger than 1, we can guarantee an even number. However, this is not always the case as shown by solutions that I computed using code.
0.00 <= 2m <= 0.00
1.12 <= 2m <= 2.00
2.00 <= 2m <= 3.12
2.74 <= 2m <= 4.00
3.40 <= 2m <= 4.74
4.00 <= 2m <= 5.40
4.55 <= 2m <= 6.00
5.06 <= 2m <= 6.55
5.54 <= 2m <= 7.06
6.00 <= 2m <= 7.54

Other presentation (but actually still the same method): first we look at the cases where $1+8n$ is a perfect square. In this case, it is the square of an odd integer, $2k+1$ say. By simplifying the equality $(1+8n) = 2k+1)^2$, we get $n=k(k+1)/2$.
This gives the idea to compare $n$ to the triangular numbers $k(k+1)/2$. Since these numbers form an increasing sequence from $0$ to infinity, there is a unique non-negative integer $k$ such that
$$\frac{k(k+1)}{2} \le n < \frac{(k+1)(k+2)}{2}.$$
Hence
$$(2k+1)^2 \le 4k^2+4k+1 = 8n+1 < 4k^2+12k+9 = (2k+3)^2.$$
Since $(2k+3)^2 \equiv 1 \mod 8$, the last strict inequality implies $8n+9 \le (2k+3)^2$.
$$2k+1 \le \sqrt{8n+1} \text{ and } \sqrt{8n+9} \le 2k+3,$$
so $2k$ is in the interval $[\sqrt{9+8n}−3,\sqrt{1+8n}−1]$.
