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If you imagine coloring the points with $x,y$ such that $\gcd(x,y) = 1$ in black, then you can consider what it would look like to color those that have $\gcd(x,y) = 2$ in red: there would be, in some sense, about $1/4$ as many of them by the matching that takes a black point like $(3,7)$ and transforms it into a red point by multiplying each factor by $2$ to get the point $(6,14)$.

If you push this geometric view/interpretation through, then you can see that there should be $1/9$ as many of the black points when looking at those pairs with $\gcd(x,y) = 3$.

Denoting by $p$ the probability that $\gcd(x,y) = 1$ (again, to make this sensible we are not really quantifying over "all" positive integers but rather those that are $\leq N$ for some large $N$).

Then we have:

$$p + \frac{1}{2^2}p + \frac{1}{3^2}p + \frac{1}{4^2}p + \cdots = 1$$

where the left hand side is the sum of the various probabilities of different gcd's, and the right hand side is, therefore, one (since the two numbers must have some gcd).

But now we end up with:

$$p = \frac{1}{\sum_{n\geq1}\frac{1}{n^2}}$$

The denominator is $\zeta(2)$, so this is the full result (modulo slight hand waving at the top) and uses relatively simple techniques. If you really want only that the density does not go to zero, then it suffices to show that the denominator converges; this is a pre-Calculus or Calculus result with a "$p$-series for $p > 1$".

For a collection of problems that scaffold in this way, you can find some wonderfully assembled ones from PCMI here.

I have put a longer writeup around the method described below in the Aperiodical's 2024 edition of "The Big Internet Math-Off," which can be found here!

If you imagine coloring the points with $x,y$ such that $\gcd(x,y) = 1$ in black, then you can consider what it would look like to color those that have $\gcd(x,y) = 2$ in red: there would be, in some sense, about $1/4$ as many of them by the matching that takes a black point like $(3,7)$ and transforms it into a red point by multiplying each factor by $2$ to get the point $(6,14)$.

If you push this geometric view/interpretation through, then you can see that there should be $1/9$ as many of the black points when looking at those pairs with $\gcd(x,y) = 3$.

Denoting by $p$ the probability that $\gcd(x,y) = 1$ (again, to make this sensible we are not really quantifying over "all" positive integers but rather those that are $\leq N$ for some large $N$).

Then we have:

$$p + \frac{1}{2^2}p + \frac{1}{3^2}p + \frac{1}{4^2}p + \cdots = 1$$

where the left hand side is the sum of the various probabilities of different gcd's, and the right hand side is, therefore, one (since the two numbers must have some gcd).

But now we end up with:

$$p = \frac{1}{\sum_{n\geq1}\frac{1}{n^2}}$$

The denominator is $\zeta(2)$, so this is the full result (modulo slight hand waving at the top) and uses relatively simple techniques. If you really want only that the density does not go to zero, then it suffices to show that the denominator converges; this is a pre-Calculus or Calculus result with a "$p$-series for $p > 1$".

For a collection of problems that scaffold in this way, you can find some wonderfully assembled ones from PCMI here.

If you imagine coloring the points with $x,y$ such that $\gcd(x,y) = 1$ in black, then you can consider what it would look like to color those that have $\gcd(x,y) = 2$ in red: there would be, in some sense, about $1/4$ as many of them by the matching that takes a black point like $(3,7)$ and transforms it into a red point by multiplying each factor by $2$ to get the point $(6,14)$.

If you push this geometric view/interpretation through, then you can see that there should be $1/9$ as many of the black points when looking at those pairs with $\gcd(x,y) = 3$.

Denoting by $p$ the probability that $\gcd(x,y) = 1$ (again, to make this sensible we are not really quantifying over "all" positive integers but rather those that are $\leq N$ for some large $N$).

Then we have:

$$p + \frac{1}{2^2}p + \frac{1}{3^2}p + \frac{1}{4^2}p + \cdots = 1$$

where the left hand side is the sum of the various probabilities of different gcd's, and the right hand side is, therefore, one (since the two numbers must have some gcd).

But now we end up with:

$$p = \frac{1}{\sum_{n\geq1}\frac{1}{n^2}}$$

The denominator is $\zeta(2)$, so this is the full result (modulo slight hand waving at the top) and uses relatively simply techniques. If you really want only that the density does not go to zero, then it suffices to show that the denominator converges; this is a pre-Calculus or Calculus result with a "$p$-series for $p > 1$".

For a collection of problems that scaffold in this way, you can find some wonderfully assembled ones from PCMI here.

If you imagine coloring the points with $x,y$ such that $\gcd(x,y) = 1$ in black, then you can consider what it would look like to color those that have $\gcd(x,y) = 2$ in red: there would be, in some sense, about $1/4$ as many of them by the matching that takes a black point like $(3,7)$ and transforms it into a red point by multiplying each factor by $2$ to get the point $(6,14)$.

If you push this geometric view/interpretation through, then you can see that there should be $1/9$ as many of the black points when looking at those pairs with $\gcd(x,y) = 3$.

Denoting by $p$ the probability that $\gcd(x,y) = 1$ (again, to make this sensible we are not really quantifying over "all" positive integers but rather those that are $\leq N$ for some large $N$).

Then we have:

$$p + \frac{1}{2^2}p + \frac{1}{3^2}p + \frac{1}{4^2}p + \cdots = 1$$

where the left hand side is the sum of the various probabilities of different gcd's, and the right hand side is, therefore, one (since the two numbers must have some gcd).

But now we end up with:

$$p = \frac{1}{\sum_{n\geq1}\frac{1}{n^2}}$$

The denominator is $\zeta(2)$, so this is the full result (modulo slight hand waving at the top) and uses relatively simple techniques. If you really want only that the density does not go to zero, then it suffices to show that the denominator converges; this is a pre-Calculus or Calculus result with a "$p$-series for $p > 1$".

For a collection of problems that scaffold in this way, you can find some wonderfully assembled ones from PCMI here.

If you imagine coloring the points with $x,y$ such that $\gcd(x,y) = 1$ in black, then you can consider what it would look like to color those that have $\gcd(x,y) = 2$ in red: there would be, in some sense, about $1/4$ as many of them by the matching that takes a black point like $(3,7)$ and transforms it into a red point by multiplying each factor by $2$ to get the point $(6,14)$.

If you push this geometric view/interpretation through, then you can see that there should be $1/9$ as many of the black points when looking at those pairs with $\gcd(x,y) = 3$.

Denoting by $p$ the probability that $\gcd(x,y) = 1$ (again, to make this sensible we are not really quantifying over "all" positive integers but rather those that are $\leq N$ for some large $N$).

Then we have:

$$p + \frac{1}{2^2}p + \frac{1}{3^2}p + \frac{1}{4^2}p + \cdots = 1$$

where the left hand side is the sum of the various probabilities of different gcd's, and the right hand side is, therefore, one (since the two numbers must have some gcd).

But now we end up with:

$$p = \frac{1}{\sum_{n\geq1}\frac{1}{n^2}}$$

The denominator is $\zeta(2)$, so this is the full result (modulo slight hand waving at the top) and uses relatively simply techniques. If you really want only that the density does not go to zero, then it suffices to show that the denominator converges; this is a pre-Calculus or Calculus result with a "$p$-series for $p > 1$.

For a collection of problems that scaffold in this way, you can find some wonderfully assembled ones from PCMI here.

If you imagine coloring the points with $x,y$ such that $\gcd(x,y) = 1$ in black, then you can consider what it would look like to color those that have $\gcd(x,y) = 2$ in red: there would be, in some sense, about $1/4$ as many of them by the matching that takes a black point like $(3,7)$ and transforms it into a red point by multiplying each factor by $2$ to get the point $(6,14)$.

If you push this geometric view/interpretation through, then you can see that there should be $1/9$ as many of the black points when looking at those pairs with $\gcd(x,y) = 3$.

Denoting by $p$ the probability that $\gcd(x,y) = 1$ (again, to make this sensible we are not really quantifying over "all" positive integers but rather those that are $\leq N$ for some large $N$).

Then we have:

$$p + \frac{1}{2^2}p + \frac{1}{3^2}p + \frac{1}{4^2}p + \cdots = 1$$

where the left hand side is the sum of the various probabilities of different gcd's, and the right hand side is, therefore, one (since the two numbers must have some gcd).

But now we end up with:

$$p = \frac{1}{\sum_{n\geq1}\frac{1}{n^2}}$$

The denominator is $\zeta(2)$, so this is the full result (modulo slight hand waving at the top) and uses relatively simply techniques. If you really want only that the density does not go to zero, then it suffices to show that the denominator converges; this is a pre-Calculus or Calculus result with a "$p$-series for $p > 1$".

For a collection of problems that scaffold in this way, you can find some wonderfully assembled ones from PCMI here.