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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

Greg Martin
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60q
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  • Hint: show that if a particular $n$ makes the left-hand endpoint exactly equal to an even integer, then $n+1$ makes the right-hand endpoint exactly equal to the next even integer. – Greg Martin Oct 23 '22 at 20:45
  • @GregMartin Sorry but I couldn't follow this line of reasoning. Because you can also have the left-hand endpoint be something like 3.4 or 4.55. – 60q Oct 23 '22 at 22:58
  • That's not relevant to the if-then statement I proposed. – Greg Martin Oct 23 '22 at 23:40

2 Answers2

1

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 $[\sqrt{9+8n}-3,\sqrt{1+8n}-1]$.

Christophe Leuridan
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  • I did not expect a solution like that. It feels like many leaps of faith hahaha, so I don't think I could have come up with it on my own. – 60q Oct 23 '22 at 23:01
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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 $[\sqrt{9+8n}−3,\sqrt{1+8n}−1]$.

Christophe Leuridan
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