3.1 Intersection of Unit Integer Circles

Recall that the density of a unit integer circle $6/\pi^{2}>1/2$ and therefore the intersection of two or fewer integer circles is non-zero. In fact, for three circles this is also true. In this section, we show that this intersection is dense in $\mathbb{Z}^{2}$ with respect to our choice of exhaustive filtrations, and compute that density.

3.1.1 Density of Two Integer Circles with respect to $\mathcal{I}$

Definition 3.1.1.

Let $m$ be a positive integer. Denote by $D(m)$ the relative density of the intersection of two integer unit circles with centres $(0,0)$ and $(m,0)$ with respect to the intersection of integer unit circles with centres $(0,0)$ and $(1,0)$ . Namely,

D(m)=\frac{\#_{\mathcal{I}}\big{(}S_{1}(0,0)\cap S_{1}(m,0)\big{)}}{\#_{% \mathcal{I}}\big{(}S_{1}(0,0)\cap S_{1}(1,0)\big{)}}.

Proposition 3.1.2.

The relative density function $D$ is multiplicative. Furthermore, for $p\in{\mathbb{P}}$ and $k\in\mathbb{N}$ we have:

D(p^{k})=\frac{p^{2}-1}{p^{2}-2}.

Theorem 3.1.3.

The density of the intersection of two integer unit circles is given by

\#_{\mathcal{I}}\big{(}S_{1}(0,0)\cap S_{1}(m,0)\big{)}=D(m)\prod_{x\in{% \mathbb{P}}}\left(1-\frac{2}{x^{2}}\right).

Example 3.1.4.

In this example, we use Python to approximate the density

\#_{\mathcal{I}}(S_{1}(0,0)\cap S_{1}(1,0)).

To see the program used for this numerical experiment, see Appendix A.

From our experiments, we see that the density is approximately 0.3219936906250593. A regular continued fraction for this number is:

0.3219936906250593=[0;3,9,2,6,1,2,31,2,1,3,2,3883].

As 31 is a large term, this means that this term (and any following terms) contribute very little to the value of the continued fraction. Hence

0.3219936906250593\approx[0;3,9,2,6,1,2]=\frac{407}{1264}.

From Theorem 3.1.3, we have this ntersection should be

\prod_{x\in{\mathbb{P}}}\left(1-\frac{2}{x^{2}}\right).

From experimentation, we have

\prod_{x\in{\mathbb{P}}}\left(1-\frac{2}{x^{2}}\right)\approx 0.32264229458178% 695.

Using the continued fraction, we find

0.32264229458178695=[0;3,10,16,1,3,1,1,2,5,23,1,1,8,7]

\approx[0;3,10,16,1,3,1,1,2,5]=\frac{20934}{64883}.

The ratio of these values is $1.0020143374718573$ , which is close to $1$ This provides basis for a guess that the two are equal, which we will now prove.

Remark 3.1.5.

The constant $p=\prod\limits_{x\in{\mathbb{P}}}(1-2/x^{2})$ is found elsewhere in mathematics.

In number theory, this constant relates to the Feller–Tornier constant . Particularly, letting $C_{FT}$ , be the Feller-Tornier constant, we have the equality:

p=2C_{FT}-1.

This equality lets us express $p$ in terms of the prime zeta function $P$ , where

P(s)=\sum_{p\in{\mathbb{P}}}\frac{1}{p^{s}}.

We write that

p=\prod\limits_{x\in{\mathbb{P}}}(1-2/x^{2})=\exp\left({\frac{-\sum_{n=1}^{% \infty}2^{n}P(2n)}{n}}\right).

This constant, along with $12/\pi^{2}=2(6/pi^{2})$ , appears in the formula for finding the density of Petersen graphs . That is

\frac{2}{\zeta(2)}-p.

Also, we have that the density of integers $n$ such that the Möbius function $\mu(n)=\mu(n+1)=0$ is given

1-\frac{2}{\zeta(2)}+p.

This is interesting to notice in comparison with this research, as this formula is the difference between twice the density of one integer circle and the density of the intersection of two integer circles.

The constant $p$ is also known to be the density of integers $n$ such that both $n$ and $n+1$ are not divisible by a prime number raised to the power of 2 (i.e. both $n$ and $n+1$ are square-free).

Example 3.1.6.

Now, we consider a few numerical examples used to approximate $D$ . We used a brute-force algorithm to find and count points of intersection. For reference of the program used, see Appendix B. We compare the count of points in the intersection of unit circles with centres $(0,0)$ and $(a,0)$ with the count of points in the intersection of unit circles with centres $(0,0)$ and $(1,0)$ . Then, we computed the continued fractions using an online tool to see if a simpler rational form would be a good approximatino of this ratio.

a	Ratio	Continued Fraction	Approximation
2	0.6665676494129102	0; 1, 1, 1, 1121, 2, 8, 2, 341108	2/3
3	0.8756286853971558	0; 1, 7, 24, 1, 2, 1, 2, 6, 1, 2, 1811397	7/8
4	0.6667877277116986	0; 1, 2, 917, 2, 10, 2, 181214	2/3
5	0.9581771599373725	0; 1, 22, 1, 10, 6, 3, 7, 1, 1106357	23/24
6	0.5833309263163976	0; 1, 1, 2, 1, 1, 2884, 1, 1, 476384	7/12

The rational approximations with smaller $p$ and $q$ agree with the formula in Proposition 3.1.2.

Remark 3.1.7.

Notice that the density of intersection of two integer unit circles is a rational coefficient of the density of the intersection of two unit circles unit distance apart.

Proof of Theorem 3.1.3

To construct a proof, we use the following lemmas:

Lemma 3.1.8.

Let $n\in\mathbb{Z}_{>0}$ . Then the following expressions are equal

\sum_{d|n}\mu(n)=[n=1].

\comment

prove this in the presentation

Proof.

First, let $n=1$ . Then the set of divisors of $n$ is $\{1\}$ . Then

\sum_{d|n}\mu(n)=\mu(1)=1=[1=1].

Now suppose $n\neq 1$ . Then $n=p_{1}^{r_{1}}\cdot p_{2}^{r_{2}}\cdots p_{k}^{r_{k}}$ . Then $n$ has set of divisors:

\{p_{1}^{l_{1}}\cdot p_{2}^{l_{2}}\cdots p_{k}^{l_{k}}:l_{i}\leq r_{i}\}.

Summing the Möbius function over these divisors gives

	$\displaystyle\sum\limits_{d\|n}\mu(n)=\,$	$\displaystyle\mu(1)+\mu(p_{1})+\ldots+\mu(p_{k})$
		$\displaystyle+\mu(p_{1}\cdot p_{2})+\ldots+\mu(p_{1}\cdot p_{k})+\ldots+\mu(p_% {k-1}\cdot p_{k})$
		$\displaystyle+\ldots+\mu(p_{1}\cdots p_{k})$
	$\displaystyle=\,$	$\displaystyle{k\choose 1}-{k\choose 2}+{k\choose 3}+\ldots+(-1)^{k}{k\choose k}$
	$\displaystyle=\,$	$\displaystyle\sum\limits_{j=0}^{k}(-1)^{j}{k\choose j}.$

Now we show that the above sum is equal to zero by induction for integers $k\geq 2$ .

Base of induction. Let $k=1$ . Then we have

\sum_{j=0}^{k}(-1)^{j}{k\choose j}=\sum_{n=0}^{1}(-1)^{n}{1\choose n}={1% \choose 0}-{1\choose 1}=1-1=0.

Step of induction. Now suppose that this holds for some $k$ that

\sum_{j=0}^{k}(-1)^{j}{k\choose j}=0.

Now recall the following recursive formula for binomial coefficients:

{a\choose b}={a-1\choose b}+{a-1\choose b-1}.

To prepare for substituting this formula in our sums, we rewrite the formula for $\sum_{d|n}\mu(n)$ as:

\sum_{j=0}^{k+1}(-1)^{j}{k+1\choose j}=\left(\sum_{j=1}^{k}(-1)^{j}{k+1\choose j% }\right)+(-1)^{0}{k+1\choose 0}+(-1)^{k+1}{k+1\choose k+1}

=\left(\sum_{j=1}^{k}(-1)^{j}{k+1\choose j}\right)+(-1)^{0}+(-1)^{k+1}=\left(% \sum_{j=1}^{k}(-1)^{j}{k+1\choose j}\right)+1+(-1)^{k+1}.

\sum_{j=1}^{k}(-1)^{j}{k+1\choose j}+1+(-1)^{k+1}=

\left(\sum_{j=1}^{k}(-1)^{j}{k\choose j}+(-1)^{j}{k\choose j-1}\right)+1+(-1)^% {k+1}

=\left(\sum_{j=1}^{k}(-1)^{j}{k\choose j}+1\right)+\sum_{j=1}^{k}(-1)^{j}{k% \choose j-1}-(-1)^{k}

=\sum_{j=0}^{k}{k\choose j}+\sum_{j=0}^{k-1}(-1)^{j+1}{k\choose j}-(-1)^{k}=% \sum_{j=0}^{k}{k\choose j}-\sum_{j=0}^{k-1}(-1)^{j}{k\choose j}-(-1)^{k}.

Notice that

\sum_{j=0}^{k-1}(-1)^{j}{k\choose j}=\sum_{j=0}^{k}(-1)^{j}{k\choose j}-(-1)^{% k}{k\choose k}=\sum_{j=0}^{k}(-1)^{j}{k\choose j}-(-1)^{k}.

Substituting this gives

\sum_{j=0}^{k+1}(-1)^{j}{k\choose j}=\sum_{j=0}^{k}{k\choose j}-\sum_{j=0}^{k-% 1}(-1)^{j}{k\choose j}-(-1)^{k}

=\sum_{j=0}^{k}{k\choose j}-\sum_{j=0}^{k}(-1)^{j}{k\choose j}+(-1)^{k}-(-1)^{% k}=2\sum_{j=0}^{k}(-1)^{j}{k\choose j}=2(0)=0.

Hence for $n\neq 1$ , $\sum_{d|n}\mu(n)=0$ , so we have

\sum_{d|n}\mu(n)=[n=1].

∎

Lemma 3.1.9.

The following holds:

{\#_{\mathcal{I}}\big{(}S_{1}(0,0)\cap S_{1}(1,0)\big{)}}=\prod_{x\in{\mathbb{% P}}}\left(1-\frac{2}{x}\right).

Remark 3.1.10.

The expression in Lemma 3.1.9 is the constant denominator of the function $D$ .

Proof.

We count the points in the intersection of unit integer circles with centers $(0,0),(1,0)$ within $[1,R]^{2}$ for some integer $R$ . This is given by the sum

\sum_{x,y\in[1,R]\cap\mathbb{Z}}[\gcd(x,y)=1]\cdot[\gcd(x-1,y)=1].

From Lemma 3.1.8, this is equal to

\sum_{x,y\in[1,R]\cap\mathbb{Z}}\left(\bigg{(}\sum_{d_{1}|\gcd(x,y)}\mu(d_{1})% \bigg{)}\cdot\bigg{(}\sum_{d_{2}|\gcd(x-1,y)}\mu(d_{2})\bigg{)}\right).

Expanding brackets:

\sum_{x,y\in[1,R]\cap\mathbb{Z}}\left(\sum_{d_{1}|\gcd(x,y)\atop{d_{2}|\gcd(x-% 1,y)}}\mu(d_{1})\mu(d_{2})\right).

We rewrite this as:

\sum_{1\leq d_{1},d_{2}\leq R}\left(\mu(d_{1})\mu(d_{2})\sum_{x,y\leq R}[d_{1}% |x,d_{2}|x-1,d_{1}|y,d_{2}|y]\right).

Since $d_{1}|x$ (ie $x\equiv 0\mod d_{1}$ ) and $d_{2}|x-1$ (ie $x\equiv 1\mod d_{2}$ ), we have $\gcd(d_{1},d_{2})=1$ . Hence, the statements $d_{1}|y$ and $d_{2}|y$ mean that the product $d_{1}d_{2}$ is also a divisor of $y$ . Hence we have

\sum_{d_{1},d_{2}\leq R}\left(\mu(d_{1})\mu(d_{2})\sum_{x,y\leq R}[x\equiv 0% \mod d_{1},x\equiv 1\mod d_{2},y\equiv 0\mod d_{1}d_{2}]\right).

From the Chinese Remainder Theorem, we have one solution for $x$ modulo $d_{1}d_{2}$ and one solution for $y$ modulo $d_{1}d_{2}$ .

\sum_{d_{1},d_{2}\leq R\atop\gcd(d_{1},d_{2})=1}\mu(d_{1})\mu(d_{2})\left(% \frac{R}{d_{1}d_{2}}+O(1)\right)\left(\frac{R}{d_{1}d_{2}}+O(1)\right).

Hence the number of intersection points is given \commentI get that this shoild be O(R). Ask Alexey maybe? $O(R\log^{2}R)\geq O(R)$ so it shouldn’t impact the proof

R^{2}\frac{\mu(d_{1})\mu(d_{2})}{d_{1}^{2}d_{2}^{2}}+O(R\log^{2}R).

Since $\mu$ is multiplicative and $\gcd(d_{1},d_{2})=1$ , this simplifies to

R^{2}\frac{\mu(d_{1}d_{2})}{d_{1}^{2}d_{2}^{2}}+O(R\log^{2}R).

To find the density, we take the limit of this as $R\to\infty$ and divide by $R^{2}$ . Hence

\#_{\mathcal{I}}(S_{1}(0,0)\cap S_{1}(1,0))=\lim_{R\to\infty}\frac{R^{2}\frac{% \mu(d_{1}d_{2})}{d_{1}^{2}d_{2}^{2}}}{R^{2}}=\sum_{d_{1},d_{2}:\gcd(d_{1},d_{2% })=1}^{\infty}\frac{\mu(d_{1}d_{2})}{d_{1}^{2}d_{2}^{2}}.

Let $d=d_{1}d_{2}$ . Then we have the density is

\sum_{d=1}^{\infty}\sum_{d_{1}d_{2}=d\atop\gcd(d_{1},d_{2})=1}\frac{\mu(d)}{d^% {2}}

=\sum_{d=1}^{\infty}\frac{\mu(d)}{d^{2}}\sum_{d_{1}d_{2}=d\atop\gcd(d_{1},d_{2% })=1}1.

The number of divisors of $d$ (ie - the number of possible $d_{1}$ ) is given by $\tau$ . However, we only count $d_{1}$ such that $\gcd(d_{1},d/d_{1})=1$ . If $\gcd(d_{1},d/d_{1})\neq 1$ then there exists prime number $p$ such that $p|d_{1}$ and $p|d/d_{1}$ , so $p^{2}|d$ . So $\mu(d)=0$ makes no contribution to the sum. We write

\sum_{d=1}^{\infty}\frac{\mu(d)\tau(d)}{d^{2}}=\prod_{p}\left(1+\frac{\mu(p)% \tau(p)}{p^{2}}+\frac{\mu(p^{2})\tau(p^{2})}{p^{4}}+\frac{\mu(p^{3})\tau(p^{3}% )}{p^{9}}+\ldots\right)

=\prod_{p}\left(1+\frac{\mu(p)\tau(p)}{p^{2}}\right)=\prod_{p}\left(1-\frac{2}% {p^{2}}\right).

∎

The statement of Theorem 3.1.3 follows from Definition 3.1.1 and Lemma 3.1.9.∎

Proof of Proposition 3.1.2

The proof for Proposition 3.1.2 will be seperated into three lemmas.

Lemma 3.1.11.

Let $p$ be a prime number. Then

D(p)=\frac{p^{2}-1}{p^{2}-1}.

Proof.

First, we will prove the given formula holds for $D(p)$ , where $p$ is a prime number.

We count the points in the intersection of unit integer circles with centers $(0,0),(1,0)$ within $[1,R]^{2}$ for some integer $R$ . This is given by the sum

\sum_{x,y\in[1,R]\cap\mathbb{Z}}[\gcd(x,y)=1]\cdot[\gcd(x-p,y)=1].

From Lemma 3.1.8, this is equal to

\sum_{x,y\in[1,R]\cap\mathbb{Z}}\left(\sum_{d_{1}|\gcd(x,y)}\mu(d_{1})\cdot% \sum_{d_{2}|\gcd(x-p,y)}\mu(d_{2})\right).

Expanding brackets, we get:

\sum_{x,y\in[1,R]\cap\mathbb{Z}}\left(\sum_{d_{1}|\gcd(x,y)}\bigg{(}\sum_{d_{2% }|\gcd(x-p,y)}\mu(d_{1})\mu(d_{2})\bigg{)}\right).

We rewrite this as:

\sum_{1\leq d_{1},d_{2}\leq R}\left(\mu(d_{1})\mu(d_{2})\cdot\bigg{(}\sum_{x,y% \leq R}[d_{1}|x,d_{2}|x-p,d_{1}|y,d_{2}|y]\bigg{)}\right).

Since $d_{1}|x$ (i.e. there exists integer $k_{1}$ such that $x=k_{1}d_{1}$ ) and $d_{2}|x-p$ (i.e. there exists integer $k_{2}$ such that $x=k_{2}d_{2}+p$ ). Then we have $p=k_{1}d_{1}-k_{2}d_{2}$ , so $\gcd(d_{1},d_{2})$ is a divisor of $p$ . Note that since $p$ is prime, we have that $\gcd(d_{1},d_{2})=1$ or $\gcd(d_{1},d_{2})=p$ .

Note that $d_{1}|y$ and $d_{2}|y$ gives $\operatorname{lcm}(d_{1}d_{2})|y$ .

\sum_{d_{1},d_{2}\leq R}\left(\mu(d_{1})\mu(d_{2})\cdot\bigg{(}\sum_{x,y\leq R% }[x\equiv 0\mod d_{1},x\equiv p\mod d_{2},y\equiv 0\mod\operatorname{lcm}(d_{1% }d_{2})]\bigg{)}\right).

Evidently, there is one solution for $y$ modulo $\operatorname{lcm}(d_{1}d_{2})$ . If $d_{1}$ and $d_{2}$ are co-prime, then by the Chinese Remainder Theorem there is one solution for $x$ modulo $d_{1}d_{2}=\operatorname{lcm}(d_{1}d_{2})$ . If $\gcd(d_{1},d_{2})=p$ , let $d_{1}=a_{1}p$ and $d_{2}=a_{2}p$ . Then there exist integers $k_{1},k_{2}$ such that

k_{1}a_{1}p=x=k_{2}a_{2}p+p.

Thus

k_{1}a_{1}=\frac{x}{p}=k_{2}a_{2}+1.

Since $\gcd(d_{1},d_{2})=p$ , $\gcd(a_{1},a_{2})=1$ , so by the Chinese Remainder Theorem we have one solution for $x/p$ modulo $a_{1}a_{2}$ . Hence there is one solution for $x$ modulo $a_{1}a_{2}p=d_{1}d_{2}/p=\operatorname{lcm}(d_{1}d_{2})$ .

We have the number of intersection to be points:

\sum_{d_{1},d_{2}\leq R,\gcd(d_{1},d_{2})|p}\mu(d_{1})\mu(d_{2})\left(\frac{R}% {\operatorname{lcm}(d_{1},d_{2})}+O(1)\right)\left(\frac{R}{\operatorname{lcm}% (d_{1},d_{2})}+O(1)\right).

Seperating this into the two cases for $\gcd(d_{1},d_{2})$ we get

\sum_{d_{1},d_{2}\leq R,\gcd(d_{1},d_{2})=1}\mu(d_{1})\mu(d_{2})\left(\frac{R}% {d_{1}d_{2}}+O(1)\right)\left(\frac{R}{d_{1}d_{2}}+O(1)\right)+

\sum_{d_{1},d_{2}\leq R,\gcd(d_{1},d_{2})=p}\mu(d_{1})\mu(d_{2})\left(\frac{Rp% }{d_{1}d_{2}}+O(1)\right)\left(\frac{Rp}{d_{1}d_{2}}+O(1)\right).

In the second term, let $d_{1}=a_{i}p_{i}$ . Then

\sum_{d_{1},d_{2}\leq R,\gcd(d_{1},d_{2})=1}\mu(d_{1})\mu(d_{2})\left(\frac{R}% {d_{1}d_{2}}+O(1)\right)\left(\frac{R}{d_{1}d_{2}}+O(1)\right)+

\sum_{a_{1},a_{2}\leq R/p,\gcd(a_{1},a_{2})=1}\mu(pa_{1})\mu(pa_{2})\left(% \frac{R}{a_{1}a_{2}p}+O(1)\right)\left(\frac{Rp}{a_{1}a_{2}p}+O(1)\right).

Since $\mu(pa_{1})=-\mu(a_{1})$ if $\gcd(a_{1},p)=1$ and 0 otherwise. Hence

\sum_{a_{1},a_{2}\leq R/p,\gcd(a_{1},a_{2})=1}\mu(pa_{1})\mu(pa_{2})\left(% \frac{R}{a_{1}a_{2}p}+O(1)\right)^{2}

=\sum_{a_{1},a_{2}\leq R/p,\gcd(a_{1},a_{2})=1,\gcd(a_{i},p)=1}\mu(a_{1})\mu(a% _{2})\left(\frac{R}{a_{1}a_{2}p}+O(1)\right)^{2}.

Rather than only counting $a_{i}$ coprime with $p$ , we count all terms and subtract the non-comprime ones afterwards:

	$\displaystyle\sum_{a_{1},a_{2}\leq R/p,\gcd(a_{1},a_{2})=1}$	$\displaystyle\mu(pa_{1})\mu(pa_{2})\left(\frac{R}{a_{1}a_{2}p}+O(1)\right)^{2}$
$\displaystyle-$	$\displaystyle\sum_{b_{1}\leq R/p^{2},a_{2}\leq R/p\gcd(b_{1},a_{2})=1}$	$\displaystyle\mu(pb_{1})\mu(a_{2})\left(\frac{R}{b_{1}a_{2}p^{2}}+O(1)\right)^% {2}$
$\displaystyle-$	$\displaystyle\sum_{a_{1}\leq R/p,b_{2}\leq R/p^{2}\gcd(a_{1},b_{2})=1}$	$\displaystyle\mu(a_{1})\mu(pb_{2})\left(\frac{R}{a_{1}b_{2}p^{2}}+O(1)\right)^% {2}.$

Let

	$\displaystyle A$	$\displaystyle=$	$\displaystyle\sum_{a_{1},a_{2}\leq R/p,\gcd(a_{1},a_{2})=1}\mu(pa_{1})\mu(pa_{% 2})\left(\frac{R}{a_{1}a_{2}p}+O(1)\right)^{2};$
	$\displaystyle B$	$\displaystyle=$	$\displaystyle\sum_{b_{1}\leq R/p^{2},a_{2}\leq R/p\gcd(b_{1},a_{2})=1}\mu(pb_{% 1})\mu(a_{2})\left(\frac{R}{b_{1}a_{2}p^{2}}+O(1)\right)^{2}.$

Then we the number of points in the intersection while $\gcd(d_{1},d_{2})=p$ is $A-2B$ , as the two subtracted terms are the same up to a change of iterative variable.

Using the same method as before, notice that

$\displaystyle B$	$\displaystyle=$	$\displaystyle\sum_{b_{1}\leq R/p^{2},a_{2}\leq R/p\gcd(b_{1},a_{2})=1}\mu(p)% \mu(b_{1})\mu(a_{2})\left(\frac{R}{b_{1}a_{2}p^{2}}+O(1)\right)^{2}$
$\displaystyle-$	$\displaystyle\sum_{c_{1}\leq R/p^{3},a_{2}\leq R/p\gcd(c_{1},a_{2})=1}\mu(p)% \mu(pc_{1})\mu(a_{2})\left(\frac{R}{c_{1}a_{2}p^{3}}+O(1)\right)^{2}$
$\displaystyle-$	$\displaystyle\sum_{\hat{a}_{1}\leq R/p^{2},b_{2}\leq R/p^{2}\gcd(c_{1},a_{2})=% 1}\mu(p)\mu(b_{1})\mu(pb_{2})\left(\frac{R}{b_{1}b_{2}p^{3}}+O(1)\right)^{2}.$

Here, $\mu(p)=-1$ . Up to a change in variable between $b$ and $c$ , the last two subtracted terms are identical. Hence,

	$\displaystyle B=$	$\displaystyle-$	$\displaystyle\sum_{b_{1}\leq R/p^{2},a_{2}\leq R/p\gcd(b_{1},a_{2})=1}\mu(b_{1% })\mu(a_{2})\left(\frac{R}{b_{1}a_{2}p^{2}}+O(1)\right)^{2}$
		$\displaystyle+$	$\displaystyle 2\sum_{c_{1}\leq R/p^{3},a_{2}\leq R/p\gcd(c_{1},a_{2})=1}\mu(pc% _{1})\mu(a_{2})\left(\frac{R}{c_{1}a_{2}p^{3}}+O(1)\right)^{2}.$

To find density, we take the limit as $R$ goes to infinity of the number of points divided by $R^{2}$ .

\lim_{R\to\infty}\frac{A}{R^{2}}=\lim_{R\to\infty}\sum_{\gcd(d_{1},d_{2})=1,d_% {1},d_{2}\leq R}\frac{\mu(d_{1})\mu(d_{2})}{d_{1}^{2}d_{2}^{2}p^{2}}+O(R\log^{% 2}R)=\frac{1}{p^{2}}\prod_{x\in{\mathbb{P}}}\left(1-\frac{2}{x}\right).

\lim_{R\to\infty}\frac{B}{R^{2}}=-\frac{1}{p^{2}}\frac{(A-2B)}{R^{2}}.

Substituting the expression for $B$ into itself repeatedly, we find the density while $\gcd(d_{1},d_{2})=p$ is:

\sum_{n=1}^{\infty}\frac{2^{n-1}}{p^{2n}}\cdot\prod_{x\in{\mathbb{P}}}\left(1-% \frac{2}{x}\right)

Thus, the density of the intersection of the circles is:

\left(1+\sum_{n=1}^{\infty}\frac{2^{n-1}}{p^{2n}}\right)\prod_{x\in{\mathbb{P}% }}\left(1-\frac{2}{x}\right)=\frac{p^{2}-1}{p^{2}-2}\prod_{x\in{\mathbb{P}}}% \left(1-\frac{2}{x}\right).

∎

Lemma 3.1.12.

Let $p$ be a prime number and $k$ a positive integer. Then

D(p^{k})=D(p)=\frac{p^{2}-1}{p^{2}-2}.

Proof.

Consider $p^{l}$ with $l>1$ . Then $\mu(p^{l})=0$ , so when $\gcd(d_{1},d_{2})=p^{l}$ we get $\mu(d_{1})=\mu(d_{2})=0$ . Hence $D(p^{k})-D(p)=0$ , so $D(p^{k})=D(p).$ ∎

Now we show that $D$ is a multiplicative function.

Lemma 3.1.13.

The relative density function $D$ is multiplicative. That is, for integers $m,n$ with no shared prime factors, we have that

D(mn)=D(m)\cdot D(n).

Proof.

Consider the integer circles with centres at $(0,0)$ and $(0,mn)$ with $\gcd(m,n)=1$ . We have

D(mn)/\#_{\mathcal{I}}(S_{1}(0,0)\cap S_{1}(1,0))=\sum_{d_{1},d_{2}:\gcd(d_{1}% ,d_{2})|mn}^{\infty}\frac{\mu(d_{1})\mu(d_{2})}{\operatorname{lcm}(d_{1}d_{2})% ^{2}}.

Since $\gcd(d_{1},d_{2})$ divides $mn$ and hence both $d_{1}$ an $d_{2}$ divide $mn$ . We write

D(mn)/\#_{\mathcal{I}}(S_{1}(0,0)\cap S_{1}(1,0))=\sum_{d_{1}|mn,d_{2}|mn}% \frac{\mu(d_{1})\mu(d_{2})}{\operatorname{lcm}(d_{1},d_{2})^{2}}.

Let $d_{1}=r_{1}s_{1},d_{2}=r_{2}s_{2}$ with $r_{i}|m$ and $s_{i}|n$ . Notice that $\gcd(r_{i},s_{i})=1$ . Then

D(mn)/\#_{\mathcal{I}}(S_{1}(0,0)\cap S_{1}(1,0))=\sum_{r_{i}|m,s_{i}|n}\frac{% \mu(r_{1}s_{1})\mu(r_{2}s_{2})}{\operatorname{lcm}(r_{1}s_{1},r_{2}s_{2})^{2}}.

Notice that $\operatorname{lcm}(r_{1}s_{1},r_{2}s_{2})=\operatorname{lcm}(r_{1},r_{2})\cdot% \operatorname{lcm}(s_{1},s_{2})$ and $\mu(r_{i}s_{i})=\mu(r_{i})\mu(s_{i})$ since $r_{i}$ and $s_{i}$ are relatively prime. Then we have

\sum_{r_{i}|m,s_{i}|n}\frac{\mu(r_{1})\mu(r_{2})\mu(s_{1})\mu(s_{2})}{% \operatorname{lcm}(r_{1},r_{2})^{2}\operatorname{lcm}(s_{1},s_{2})^{2}}=\sum_{% r_{i}|m,}\frac{\mu(r_{1})\mu(r_{2})}{\operatorname{lcm}(r_{1},r_{2})^{2}}\cdot% \sum_{s_{i}|n}\frac{\mu(s_{1})\mu(s_{2})}{\operatorname{lcm}(s_{1},s_{2})^{2}}=.

Hence

D(mn)=D(m)\cdot D(n).

∎

Combined, the statements of Lemmas 3.1.11, 3.1.12 and 3.1.13 give the statement of the proposition. ∎

Example 3.1.14.

The frequency of points in the intersection at a given y value, for x values $0\leq x<y$ is given as

f(y)=f\left(\prod p_{i}^{r_{i}}\right)=\prod p_{i}^{r_{i}-1}(p_{i}-2).

3.1.2 Density of Intersection of Three Integer Circles

A natural extension of the previous subsection is to ask about the intersection of three integer circles. We consider a numerical example. Our methods are more clearly outlined in Appendix C.

Example 3.1.15.

In this example, we consider the density of integer unit circles with centres $(0,0)$ , $(-1,0)$ and $(1,0)$ . From our Python approximation, we find

\#_{\mathcal{I}}\big{(}S_{1}(-1,0)\cap S_{1}(0,0)\cap S_{1}(1,0)\big{)}\approx 0% .2510066561587017.

By considering continued fractions, observe

0.2510066561587017=[0;3,1,61,2,1,31,1,16,12,23,1]\approx[0;3,1]=\frac{1}{4}.

We compared this number to some formulae derived from the one we found in Theorem 3.1.3. Only one such experiment produced a result of potential interest. In particular we found:

\prod_{p\in{\mathbb{P}}}\left(1-\frac{3}{p^{2}}\right)\approx 0.12549176241980% 63.

Using continued fractions, notice:

0.1254917624198063=[0;7,1,30,1,8,1,5,1,1,1,5,1,1,1,4,1,4,2]\approx[0;7,1]=% \frac{1}{8}.

The experiment provides the following guess:

\#_{\mathcal{I}}\big{(}S_{1}(-1,0)\cap S_{1}(0,0)\cap S_{1}(1,0)\big{)}=2\prod% _{p\in{\mathbb{P}}}\left(1-\frac{3}{p^{2}}\right).

This could lead to a potential generic formula for intersection of unit integer circles on a line, with distance 1 between their centres.

Example 3.1.16.

Now, consider the density

\#_{\mathcal{I}}\big{(}S_{1}(0,0)\cap S_{1}(1,0)\cap S_{1}(0,1)\big{)}.

From Python experimentation (for more details, see Appendix D), we find that

\#_{\mathcal{I}}\big{(}S_{1}(0,0)\cap S_{1}(1,0)\cap S_{1}(0,1)\big{)}\approx 0% .12548190236470805.

Using the same numerical approximation as in Example 3.1.15, observe that

\frac{0.12548190236470805}{\prod\limits_{p\in{\mathbb{P}}}\left(1-\frac{3}{p^{% 2}}\right)}\approx 0.999921428666646=[0;1,12726,3,2,7,1,17,1,5].

This number is very close to 1, so we guess that

\#_{\mathcal{I}}\big{(}S_{1}(0,0)\cap S_{1}(1,0)\cap S_{1}(0,1)\big{)}=\prod% \limits_{p\in{\mathbb{P}}}\left(1-\frac{3}{p^{2}}\right).

3.1.3 Density of Intersection with respect to $\mathcal{B}$

In this subsection, we investigate the density of the same sets as studied before with respect to a different exhaustive filtration. Specifically, we study with respect to the butterfly filtration generated by $(1,1)$ .

Counting the intersection of integer circles at a given horizontal slice

This method for counting the intersection of circles, we will consider the points $(x,y)$ on the intersection of unit integer circles for a given value $y$ .

Definition 3.1.17.

A horizontal-slice counting function $C_{s}$ is a multiplicative function belonging to the family defined by the following rule:

C_{k}(p^{m})=p^{m-1}(p-k).

Theorem 3.1.18.

Consider integer unit circles with centres $(a_{1},0)$ , $(a_{2},0)\ldots(a_{n},0)$ . Then the number of intersection points of these integer circles on a horizontal slice $y=const$ with $0\leq x<y$ is given by $C_{k}$ where $k$ is the size of the set $\{[a_{1}]_{p},\ldots,[a_{n}]_{p}\}$ .

Proof.

First we will consider the case where $m=1$ .

For a prime $p$ , if $\gcd(a,p)\not=1$ then $a\equiv 0\mod p$ .

If a point $(x,p)$ is in the intersection of the circles then

\gcd(x-a_{1},p)=\gcd(x-a_{2},p)\ldots=\gcd(x-a_{n},p)=1.

To find the excluded points, we look for points when at least one of $x-a_{j}\equiv 0\mod p.$
As we are dealing with remainders modulo $p$ , we can reduce every element modulo $p$ , so instead of having $n$ different cases, we only deal with $k$ different cases, since we only have $k$ distinct values for $a_{j}$ modulo $p$ .
Hence, for values $0\leq x<p$ we have $k$ excluded points, so the number of elements in the intersection for this range of $x$ is given as

f(p)=p-k.

For $m\neq 1$ , the exclusion is still going to be given when $x-a_{j}\equiv 0\mod p$ , but we are interested in the number of unique solutions modulo $p^{m}$ .
Since $p^{m}=(p^{m-1})p$ , we see that we have $p^{m-1}$ as many solutions the slice at (which slice). Hence, we see that $f(p^{m})=C_{k}(p^{m})$ .

Now let us show that $f$ is a multiplicative, which will then show that $f(y)=C_{k}(y)$ for $y\neq p^{m}$ . We will use the Chinese Remainder Theorem to do this.

Let $m,n$ be two coprime positive integers. Now let us consider the intersection of integer circles and the line $y=mn$ . Recall that if $\gcd(x,mn)=1$ , then $\gcd(x,m)=1$ and $\gcd(x,n)=1$ .

When we count the intersection of circles on row $mn$ , we have that $\gcd(x-a_{i},mn)=1$ . This means that $\gcd(x-a_{i},m)=\gcd(x-a_{i},n)=1.$

Considering remainders modulo $mn$ , we have $nf(m)$ solutions for $\gcd(x-a_{i},m)=1$ and $mf(n)$ solutions for $\gcd(x-a_{i},n)=1$ . Using the Chinese Remainder Theorem, we get the number of solutions to these taken simultaneously is given

F(mn)=f(m)f(n).

Hence $f$ is multiplicative, so $f(mn)=C_{k}(mn)$ . ∎

Notice that for $y>a_{n}$ , the number of residues modulo $y$ is constant. Particuarly, we have $k=n$ .

Corollary 3.1.19.

In case of convergency, the density of the intersection of unit integer circles with centres on the line $y=0$ is given by the following

\#_{\mathcal{B}(1,1)}\big{(}S_{1}(0,0)\cap S_{1}(0,a_{1})\cap\ldots\cap S_{1}(% 0,a_{k})\big{)}=\lim_{j\to\infty}\left(\frac{2}{j^{2}+j}\sum_{y=1}^{j}C_{n}(y)% \right).

Proof.

Consider the filtration $\mathcal{B}(1,1)=(B_{n})$ . Due do integer congruence, the proportion of points in the intersection of integer circles in $B_{n}$ is equal to the proportion of points in the triangle between the line $y=x$ , $y=n$ and $x=0$ .

The number of intersection points for $(x,y)$ with $0\leq x<y$ for constant $y$ is given as $C_{k}(y)$ .

Clearly, the number of intersection points for $(x,y)$ with $0\leq x<y$ and $0\leq y<k$ is the sum of $f$ from $1$ to $k$ , and considering the origin, which is not included in our circle. So we have the frequency given as

\sum_{y=1}^{k}C_{n}(y).

To find the density, we divide the frequency by the size of the larger subset $\mathcal{B}$ .

The number of lattice points for $(x,y)$ with $0\leq x<y$ for constant $y$ is clearly equal to $y$ . So, the number of lattice points for $1\leq y<k$ is given as

\sum_{y=1}^{k}y=\frac{k^{2}+k}{2}.

Dividing the number of intersection points by the number of lattice points gives that the density is

\lim_{k\to\infty}\frac{2\sum_{y=1}^{k}C_{n}(y)}{k^{2}+k}.

∎

3.1.4 Comparison between $\#_{\mathcal{I}}$ and $\#_{\mathcal{B}(1,1)}$

Example 3.1.20.

In this example, let $S$ be the intersection of unit integer circles with centres $(0,0)$ and $(1,0)$ .
Using numerical approximations, we compared the density of the intersection of integer circles centred at $(0,0)$ and $(1,0)$ with respect to the exhaustive filtrations $\mathcal{I}$ and $\mathcal{B}(1,1)$ . We found that

\frac{\#_{\mathcal{I}}S}{\#_{\mathcal{B}(1,1)S}}\approx 0.9977758007117438=[0;% 1,448,1,1,1,1,868827846].

The third rerm, 448. is quite large, so we can remove it from the continued fraction as a guess that it comes from approximation error. Then

\frac{\#_{\mathcal{I}}S}{\#_{\mathcal{B}(1,1)S}}\approx[0;1]=1;

3.1 Intersection of Unit Integer Circles

3.1.1 Density of Two Integer Circles with respect to ℐ\mathcal{I}

Definition 3.1.1.

Proposition 3.1.2.

Theorem 3.1.3.

Example 3.1.4.

Remark 3.1.5.

Example 3.1.6.

Remark 3.1.7.

Proof of Theorem 3.1.3

Lemma 3.1.8.

Proof.

Lemma 3.1.9.

Remark 3.1.10.

Proof.

Proof of Proposition 3.1.2

Lemma 3.1.11.

Proof.

Lemma 3.1.12.

Proof.

Lemma 3.1.13.

Proof.

Example 3.1.14.

3.1.2 Density of Intersection of Three Integer Circles

Example 3.1.15.

Example 3.1.16.

3.1.3 Density of Intersection with respect to ℬ\mathcal{B}

Counting the intersection of integer circles at a given horizontal slice

Definition 3.1.17.

Theorem 3.1.18.

Proof.

Corollary 3.1.19.

Proof.

3.1.4 Comparison between #ℐ\#_{\mathcal{I}} and #ℬ⁢(1,1)\#_{\mathcal{B}(1,1)}

Example 3.1.20.

3.1.1 Density of Two Integer Circles with respect to $\mathcal{I}$

3.1.3 Density of Intersection with respect to $\mathcal{B}$

3.1.4 Comparison between $\#_{\mathcal{I}}$ and $\#_{\mathcal{B}(1,1)}$