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 :)
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.