Is the set $\ \left\{ \frac{b-a }{c-a}:\ (a,b,c)\ \text{is a primitive Phythagorean triple with}\ a<b<c\ \right\}\ $ dense in $\ [0,1]\ ?$

Question

Is the set $\ \left\{ \frac{b-a }{c-a}:\ (a,b,c)\ \text{is a primitive Pythagorean triple with}\ a<b<c\ \right\}\ $ dense in $\ [0,1]\ $ and how do you show this?

It seems likely true based on glancing at the tree in the picture, however I am not very knowledgeable on the properties of Primitive Pythagorean triples...

Answer 1

Let me try an argument that is, perhaps, excessively geometric.

I start with the rational parametrization of the unit circle, \begin{align} x&=\frac{2t}{t^2+1}\\ y&=\frac{t^2-1}{t^2+1}\,. \end{align} You probably have seen this, maybe in a different form. But it’s clear that every rational value of $t$ gives a point on the unit circle whose both coordinates are rational, and equally, every rational point on the circle comes from a rational $t$, via $(x,y)\mapsto\frac{y+1}{x}$. For instance, $t=2$ corresponds to the point $(\frac45,\frac35)$.

And I hope you see at a glance that the (primitive) Pythagorean triangles are in one-to-one correspondence with the first-quadrant rational points on the circle.

Now, the points below the line $y=x$ for $1\le t\le1+\sqrt2$ can give us $a=(t^2-1)/(t^2+1)$, $b=2t/(t^2-1)$, and $a<b<c=1$, and if we calculate your ratio $$ R(t)=\frac{b-a}{c-a}=\frac{1+2t-t^2}2\,, $$ in which $t=1$ gives $R(1)=1$, while $R(1+\sqrt2)=0$. Now in the range $t\in[1,1+\sqrt2]$, rational values of $t$ give rational values of $R(t)$, and though these are not the only rational values of $R$, at least they are dense among the values of $R$. And thus we’re done.

Answer 2

A Euclid's formula variation lets us easily see what the differences are that go into $\dfrac{B-A}{C-A}\space$ and how $\space B-A>0 \space$ requires that $\space 2k^2>m.\space$

\begin{align*} A&=\bigg(\big(2n-1+k\big)^2-k^2\bigg)\\ &\qquad\qquad=(2n-1)^2+2(2n-1)k\\ B&=\quad 2\big(2n-1+k\big)k\\ &\qquad\qquad\qquad\qquad= 2(2n-1)k+2k^2\\ C&=\bigg(\big(2n-1+k\big)^2+k^2\bigg)\\ &\qquad\qquad=(2n-1)^2+2(2n-1)k+2k^2 \end{align*}

Examples, using this formula, are \begin{align*} &(n,2k^2)=(1,2)\quad F(n,k)=F(1,1)= (3,4,5)\\ &(n,2k^2)=(2,18)\quad F(n,k)=F(2,3)= (27,36,45)\\ &(n,2k^2)=(3,32)\quad F(n,k)=F(3,4)= (65,72,97)\\ &(n,2k^2)=(4,50)\quad F(n,k)=F(4,5)= (119,120,169)\\ \end{align*} The formula produces non-trivial triples for all pairs of natural numbers.

\begin{equation} n,k\in\mathbb{N} \implies\\ A=\big\{3,5,7,\cdots\big\}\\ B=\big\{4,8,12,\cdots\big\}\\ \qquad C\subset\big\{5,9,13,\cdots,4x+1\big\} \end{equation}

$ (B-A)=2k^2-(2n-1)^2\implies \\ (B-A)\in\big\{1,7,17,23,31,\cdots\}\implies \\ (B-A)\equiv\pm 1 \pmod 8 $

$ (C-A)=2k^2\implies\\ (C-A)\in\big\{2,8,18,32,50,\cdots\big\} $

$\dfrac{B-A}{C-A} =\dfrac{2k^2-(2n -1)^2}{2k^2} =1-\dfrac{(2n-1)^2}{2k^2}$

$(B>A)\implies 2k^2 \ge (2n-1)^2\\ \implies k \ge n\\ \implies 0 < \dfrac{B-A}{C-A} < 1\\$

Note that $\space\ 2k^2 \ge (2n-1)^2\space$ is not sufficient for $\space B>A\space $ as in: $\qquad\qquad (n,2k^2)=(2,8)\quad F(n,k)=F(2,2)= (21,20,29)\\ $

We can see, with arbitrarily skinny or fat Pythagorean triples like $$F(1,5000)=(10001,50010000,50010001)\quad (C-B)=1\\ \text{ or } \\ F(23661,33461)=(5406093003,5406093004,7645370045) \\ (B-A)=1)$$ from this formula, there are infinite triples where $\space \dfrac{B-A}{C-A}\space $ is equal to or comes arbitrarily close to any number in $\space[0,1].$

Answer 3

Consider the function $$f(x) = 1-\frac{2}{x^2-2x+1}$$

This function is continuous, maps the interval $(1+\sqrt{2},\infty)$ onto $(0,1)$ and it's increasing.

Choose any interval $(u,v) \subseteq [0,1]$, the preimage of $(u,v)$ under $f$ is the interval $(f^{-1}(u),f^{-1}(v))$. Choose a rational $r/s \in (f^{-1}(u),f^{-1}(v))$ with $r$ and $s$ odd and coprime. (It's known that $\Bbb Q$ is dense in $\Bbb R$, but it's not hard to see that if you restrict to rationals with odd numerators and denominators it's still dense).

Then $f(r/s) \in (u,v)$.

But $$f(r/s) = 1- \frac{2}{(r/s)^2-2 r/s + 1}= \frac{\frac{r^2-s^2}{2}-rs}{\frac{r^2+s^2}{2}-r s} = \frac{b - a}{c-a}$$

where we defined $a=rs$, $b=\frac{r^2-s^2}{2}$ and $c=\frac{r^2+s^2}{2}$.

The triple $(a,b,c)$ is a primitive Pythagorean triple with $a<b<c$ and $\frac{b-a}{c-a} \in (u,v)$. Since $(u,v)$ was any interval contained in $[0,1]$, the set is dense.