0
$\begingroup$

Find the unique cases when ${t}^{2} - 4$ is a perfect square say, ${n}^{2}$, with height bound $|t| \le N$ for positive integer $N \ge 1$, when $t$ is a rational where $t = p/q$ and integers $p$ an $q$ are relatively prime, $|p| \le N$ and $1 \le q \le N$. I am looking for a counting like solution of the unique cases, exact if possible or more likely an asymptotic expansion.

For a given $N$ there are $N \left({2\, N + 1}\right)$ cases to consider. The relatively prime condition reduces this by $6/{\pi}^{2}$ which is the probability that two random integer are relatively prime. I left off the next order term. For the integer case the only solutions are for $t = \pm 2$ which is covered in my other calculations.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

From $t^2-4=s^2$ we get $$ t^2-s^2=4~~ \Longrightarrow ~~ (t+s)(t-s) = 4 $$ hence the general rational solution $(t,s)$ is, putting $2\lambda = t+s$: $$ \left( \lambda+\frac{1}{\lambda}, \lambda-\frac{1}{\lambda} \right). $$ (It is easy to check that this indeed solves your equation.) So now we need to find the height of $t = \lambda + 1/\lambda$ in terms of the height of $\lambda=:n/d$. We have $$ t = \lambda+\frac{1}{\lambda} = \frac{d^2+n^2}{dn}, $$ where the last fraction is obviously in lowest terms, at least if $n/d$ was. Hence the height of $t$ satisfies $$ H(t) \sim H(\lambda)^2 $$ asymptotically. Since there are $O(N^2)$ rational numbers $\lambda$ of height $< N$, the number of $t$ with $H(t)<N$ that satisfies $t^2-4=\square$ is $O(N)$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – Lorenz H Menke
    Commented Jan 27, 2016 at 2:01
  • $\begingroup$ 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 $3N / \pi^2 < f(N) < 6N / \pi^2$. $\endgroup$
    – R.P.
    Commented Jan 27, 2016 at 2:27
  • $\begingroup$ Incidentally, $0.4772$ seems close to $3/(2 \pi)$. But that could be just a coincidence. $\endgroup$
    – R.P.
    Commented Jan 27, 2016 at 2:39
  • $\begingroup$ 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 $3N/(2 \pi)$. $\endgroup$
    – R.P.
    Commented Jan 27, 2016 at 3:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .