Is there any formula to find number of Pythagorean triplets between two integers 2 and j, j>2?

Question

Given $j \geq 5$, is there a formula for the number of Pythagorean triplets $(a, b, c)$ satisfying the constraint that $a, b, c \leq j$?

There exists at least one Pythagorean triplet for $j\geq5$; the question is how to find the exact number of Pythagorean triplets for large $j$.

Answer 1

The question here was studied (in a slightly more generalized version) by Sierpinski (1906) in Polish; asymptotics were found by Fricker (1977, 1982) and Fischer (1979) both in German; and an unconditional upper bound was established by Stronina (1969) in Russian.

As for English, you can find these references (and others) in, for example, the paper:

Nowak, W. G., & Recknagel, W. (1989). The distribution of Pythagorean triples and a three-dimensional divisor problem. Math. J. Okayama Univ, 31, 213-220. Link (no paywall).

Here is an excerpt from the first page:

The reference here comes from backtracking; first:

Benito, M., & Varona, J. L. (2002). Pythagorean triangles with legs less than n. Journal of computational and applied mathematics, 143(1), 117-126. Link (no paywall).

In there, the authors point to an earlier paper (p. 118):

Kühleitner, M. (1993, December). An omega theorem on Pythagorean triples. In Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg (Vol. 63, No. 1, pp. 105-113). Springer Berlin/Heidelberg. Link.

The last paper is, unfortunately, only visible as a preview; but it is written in English (despite the German title) and contains a reference to the Nowak and Recknagel paper excerpted above.

Answer 2

The problem is equivalent to asking for all Pythagorean triples with bounds on the hypotenuse and a variant of that problem is treated in the article ENUMERATION OF ALL PRIMITIVE PYTHAGOREAN TRIPLES WITH HYPOTENUSE LESS THAN OR EQUAL TO N

A simple estimate for the number of primitive Pythagorean triples is due to Lehmer: it is less than $N/2\pi$

Turning to the original problem of enumerating all triples, the following observation helps:

$a^2+b^2=c^2 \implies a=2uv, b=v^2-u^2, c=u^2+v^2; u\le v$
which leads to the problem of finding $\frac{1}{8}\sum_{i=1}^{n}r(i)$, where $r(i)$ is the number of ways to write $i$ as the sum of two squares of signed integers (cf e.g. wolfram ).

The answer is then the aforementioned sum $\frac{1}{8}\sum_{i=1}^{n}r(i)$

Answer 3

It is an old question but I recently solved a programming problem regarding this question, the article was shared above but the method was not explicitly shared here.

Following this article: Pythagorean triangles with legs less than n there is quite a nice solution

We know that each primitive Pythagorean triple can be represented by \begin{align*} a &= x^2 - y^2 \\ b &= 2xy \\ c &= x^2 + y^2 \end{align*} Following the paper lets define a few sets. \begin{align*} R(n) &= \{(x, y) \in \mathbb{Z} \times \mathbb{Z} \mid x^2 + y^2 \leq n, 0 < x < y\}\\ P(n) &= \{(x, y) \in R(n) \mid \gcd(x, y) = 1, x + y \equiv 1 \pmod{2}\} \\ Q(n) &= \{(x, y) \in R(n) \mid \gcd(x, y) = 1\} \end{align*} Note that $|P(n)|$ is exactly the amount of Pythagorean triples with hypotenuse less than or equal to $n$, which is what we are trying to find. Now according to the paper Lemma 3 states \begin{equation*} |P(n)| = \sum_{k \geq 0} (-1)^k|Q(\frac{n}{2^k})| \end{equation*} Note that we changed the region but the proof is the same.

Now all that is left is to calculate $Q(n)$. I define a few equations which are identical to the article, but easier to read. \begin{align*} L(t) &= |\{(x, y) \in R(t^2)\}| = |R(t^2)| \\ L'(t) &= |\{(x, y) \in R(t^2) \mid \gcd(x, y) = 1\}| = |Q(t^2)| \end{align*} Following the exact same proof as the article we have \begin{align*} L(t) &= \sum_{1 \leq d \leq t} L'(\frac{t}{d}) \\ L'(t) &= \sum_{1 \leq d \leq t} \mu(d) L(\frac{t}{d}) \end{align*} Therefore we have, \begin{align*} |Q(n)| &= L'(\sqrt{n}) \\ &= \sum_{1 \leq d \leq \sqrt{n}} \mu(d) L(\frac{\sqrt{n}}{d}) \\ &= \sum_{1 \leq d \leq \sqrt{n}} \mu(d) |R(\frac{n}{d^2})| \end{align*}

Now, using brute force a computer can easily find all the lattice points in $R(n)$, and with a good Möbius sieve we can get a nice easy solution.

I have a python package where I add functions like these if anyone is interested to see the code.

1. Mobius Sieve
2. Number of primitve Pythagorean triples with hypotenuse less than n