What proportion of natural numbers are of the form $(n^2-1)/24$ for some $n\in\Bbb N$?

Question

Motivation:

I found the following theorem online recently.

Theorem: For any prime $p\ge 5$, we have $$24\mid p^2-1.$$

Proof: Let $p\ge 5$ be prime and $P=p^2-1$. Observe that $P=(p-1)(p+1)$. Since $p$ is odd, both $p-1$ and $p+1$ are even, and one is a multiple of four; hence $8\mid P$. Also, $p>3$, so $3\nmid p$, and exactly one of any three consecutive integers is a multiple of three. Considering $p-1, p, p+1$, then, we must have $3\mid P$. Hence $3\times 8=24\mid P$. $\square$

I showed this to a friend and she half-joked that this would make primes easier to find; just look for numbers of the form $(n^2-1)/24$ for natural $n$ to narrow things down. I remarked that that was kind of like saying, okay, each prime greater than two is odd, so that makes primes easier to find.

But it got me thinking . . .

The Question:

What proportion of natural numbers are of the form $$m=\frac{n^2-1}{24}$$ for $n\in\Bbb N$?

Context:

It's been a while since I studied analytic number theory, so I don't have very sophisticated thoughts on the matter.

One way to approach this, I suppose, is to study the asymptotics of $(n^2-1)/24$ as $n\to \infty$. So I guess we could look at $O(n^2)$ as $n\to \infty$. Since $0\%$ of natural numbers are perfect squares, it seems that $0\%$ of natural numbers are of the form $(n^2-1)/24$.

Why am I not convinced by this heuristic?

Well, like I said, my analytic number theory is rusty. Is "percentage" even an accurate way of describing the proportion?

Edit: I guess what I'm looking for is the natural density of such numbers.

Please help :)

Wouldn't the natural density of these numbers being zero prove the natural density of primes is zero? If so, this seems like a much simpler proof than what is known! — Shaun, May 24 '23 at 18:56
$24|m^2-1$ for all $m$ not divisible by $2$ or $3$, not just for prime $m$ — J. W. Tanner, May 24 '23 at 19:07
Generally, you search primes in a specific range. For example, all primes $p$ with 200 digits. so you should search for density of perfect squares in a specific range. — Lourrran, May 24 '23 at 19:37
Natural density of these numbers would be zero as the asymptotic will be proportional to $\frac{n}{\sqrt{ \ln n}}$ as $n \rightarrow \infty$ — MokutekiJ, May 26 '23 at 17:29
I invite you to elaborate on that in an answer, @MokutekiJ; I'm likely to accept it! — Shaun, May 26 '23 at 17:48
Have a look at this question https://math.stackexchange.com/questions/4449391/is-there-a-way-to-evaluate-de-natural-density-of-numbers-of-the-form-ax2-by2
$24(n^2-1)$ is of the form $ax^2 + by^2$ with $a=24$, $b=24$, $x = n$ and $y=1$ — MokutekiJ, May 26 '23 at 17:53
To fully understand, maybe take a look at Gauss compositions and natural densities of binary quadratic forms — MokutekiJ, May 26 '23 at 17:58
This might also be interesting: https://math.stackexchange.com/questions/4320361/density-of-integers-represented-by-a-binary-quadratic-form-is-zero/4320845#4320845 — MokutekiJ, May 26 '23 at 18:02
This goes more into depth of the constant of proportionality for the asymptotic: https://arxiv.org/pdf/1612.08307.pdf
I hope that helps ;) — MokutekiJ, May 26 '23 at 18:05
Please explain why you think that integers of the form $24(n^2-1)$ are related to the first Theorem (possibly there is a misunderstanding). — Bill Dubuque, May 27 '23 at 15:19
I told you my number theory was rusty! I mean $$\frac{n^2-1}{24},$$ of course! Never mind: I got an answer now. Thank you for asking for clarification, @BillDubuque. — Shaun, May 27 '23 at 18:24
You shouldn't accept an answer to a question that was not the one you meant. This will end up confusing many readers. — Bill Dubuque, May 27 '23 at 18:56
Okay, @BillDubuque. I have edited the question yet again. I'm sorry for the confusion. — Shaun, May 27 '23 at 19:03

score 5 · Answer 1 · answered May 27 '23 at 15:27

5

The natural density is zero, as you say. But the algebra is not at all complicated: if we set a limit $N$, then the number of terms $\le N$ is exactly the integral part of $$\sqrt{\dfrac{N}{24}+1}$$ So the density on $[1,N]$ is about $\dfrac{1}{\sqrt{24N}}$. Which tends to zero.

answered May 27 '23 at 15:27

TonyK

64,559

1

But this clearly is not the question the OP meant - see my comment on the question (and the edit to the question). – Bill Dubuque May 27 '23 at 18:55
1

Hopefully, this is a quick fix for TonyK, @BillDubuque. I don't see anything wrong with an edit to this answer. – Shaun May 27 '23 at 19:12
@BillDubuque: I answered the question as it was posted. I'm not going to go chasing after it as it changes, especially for such a trivial change. – TonyK May 28 '23 at 20:05
It won't change again, @TonyK. – Shaun May 28 '23 at 20:07
It is not respectful to the community to leave wrong answers with no warning. Why not simply make the easy update? – Bill Dubuque May 28 '23 at 20:53
@BillDubuque: You have my permission to do it yourself, if you're so bothered by it. Not respectful? Oh for heaven's sake. – TonyK May 28 '23 at 22:05

What proportion of natural numbers are of the form $(n^2-1)/24$ for some $n\in\Bbb N$?

Motivation:

The Question:

Context:

1 Answers1