Timeline for Find the rational cases where ${t}^{2} - 4$ is a perfect square with height bound $|t| \le N$ for positive integer $N \ge 1$
Current License: CC BY-SA 3.0
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| when toggle format | what | by | license | comment | |
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| Jan 27, 2016 at 3:02 | comment | added | R.P. | In fact, yes, the leading constant seems indeed to be $3/(2 \pi)$. Given $N>0$, it is easy to see that there are $\sim N \pi/2 $ pairs $(d,n)$, with $d,n$ integers and $n>0$, such that $H(t) = d^2+n^2 < N$. (The relevant pairs $(d,n)$ describe the interior of a semicircle with radius $\sqrt{N}$ in the $(d,n)$-plane.) Arguing heuristically, taking $6/\pi^2$ as the density of coprime pairs $(d,n)$, we get $\sim N \pi /2 \cdot 6/\pi^2 = 3N /\pi$ pairs $(d,n)$ that give rise to values $t$ with $H(t)<N$. Since $(d,n)$ and $(n,d)$ yield the same $t$, the count of all $t$ comes to $3N/(2 \pi)$. | |
| Jan 27, 2016 at 2:39 | comment | added | R.P. | Incidentally, $0.4772$ seems close to $3/(2 \pi)$. But that could be just a coincidence. | |
| Jan 27, 2016 at 2:27 | comment | added | R.P. | I think it should be possible to work out the leading constant using the formula for $t$ given above, but I do not have the time to do the calculation right at this moment. But from the formula $t=(d^2+n^2)/(dn)$ we get $H(\lambda)^2=\max(\left| d \right|,\left| n \right| )^2 \leq d^2+n^2 = H(t) \leq \max( 2d^2, 2n^2 ) = 2H(\lambda)^2$. Since IIRC we have $\# \{ \lambda \in \mathbb{Q} : H(\lambda) < N \} \sim 6N^2 / \pi^2$, we get for $f(N) := \# \{ t \in \mathbb{Q} : \textrm{$H(t)<N$ and $t^2-4$ is square} \}$ that $3N / \pi^2 < f(N) < 6N / \pi^2$. | |
| Jan 27, 2016 at 2:06 | vote | accept | Lorenz H Menke | ||
| Jan 27, 2016 at 2:01 | comment | added | Lorenz H Menke | This is very nice, I did not think of these simple steps. I have from building numerical tables approximately $0.4772\, N$ as the asymptotic. So I suspected that the solution was $O(N)$. All that I need is to estimate the leading coefficient plus error if possible. | |
| Jan 27, 2016 at 0:45 | history | answered | R.P. | CC BY-SA 3.0 |