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

Let $x$ be the integer part of $(\sqrt{1+8n}-1)/2$. Then
$$(\sqrt{1+8n}-1)/2 - 1 < x \le (\sqrt{1+8n}-1)/2.$$
$$\sqrt{1+8n}-3 < 2x \le \sqrt{1+8n}-1.$$
The left inequality may be written $1+8n < (2x+3)^2 = 4x(x+3)+9$. Noticing that $x(X+3)$ is always even, one sees that $(2x+3)^2 \equiv 1 \mod 8$. So the inequality $1+8n < (2x+3)^2$ implies $9+8n < (2x+3)^2$, which gives the inequality $\sqrt{9+8n}-3 \le 2x$.
Hence, the even integer $2x$ is in the interval $[\sqrt{9+8n}-3,\sqrt{1+8n}-1]$.
