A Rational Parameterization of Multiple Simple Expressions (Or the intersection of two rational parameterizations)

Question

Context

I am interested specifically in all rational values of $x$ for which $\sqrt{1-x}$ and $\sqrt{1+x}$ are rational.

In general; however, I am curious if there is a method for taking any number of expressions in the form $\sqrt{n_i \pm x}$ and finding all rational values for x that ensure all of the expressions are rational.

What I have tried

I have tried a few different methods:

Method 1

I have tried calculating rational parameterizations of both expressions individually by taking the rational point $(0,1)$ and finding the intersection between a line with a rational slope through that point and the graphs individually.

For $y=\sqrt{1-x}$ I used the line $x=t(y-1)$

Plugging this into $y=\sqrt{1-x}$ I get $y=\sqrt{1-t(y-1)}$

By solving this for $y$ (and ultimate $x$) in terms of $t$, I get $x=-t^2-2t$

Similarly for $y=\sqrt{1+x}$ with this method I get $x=t^2+2t$

But I have not been able to parameterize the intersection of these two parameterizations.

Method 2

I have tried calculating rational parameterizations of both expressions together by combining them into the system:

$$y=\sqrt{1-x}$$

$$z=\sqrt{1+x}$$

and taking the rational point $(0,1,1)$ and finding the intersection between a line with a rational slope through that point and the surface $(x,\sqrt{1-x},\sqrt{1+x})$.

By this method I get a long and messy formula that does not guarantee rational coordinates.

Method 3

I have tried using the same techniques to rationally parameterize $y=\sqrt{1-x}+\sqrt{1+x}$ with a similarly messy result.

Method 4

Since $x$ is rational, $x=\frac{a}{b}$ where $a$ and $b$ are co-prime integers, the expressions above can be rewritten as:

$$\sqrt{1-x}=\sqrt{1-\frac{a}{b}}=\sqrt{\frac{b-a}{b}}$$

$$\sqrt{1+x}=\sqrt{1+\frac{a}{b}}=\sqrt{\frac{b+a}{b}}$$

For these to be rational either $a$ must contain $b$ as a factor (which is impossible because $a$ and $b$ are defined co-prime) or $b$ must be a square integer $b=c^2$.

I performed a search for all positive integers in certain range that where $\sqrt{c^2-a}$ and $\sqrt{c^2+a}$ are integers to attempt to identify a pattern.

The first few fully reduced fractions (where $a \neq 0$) that I found are:

$$\frac{24}{25},\frac{120}{169},\frac{240}{289},\frac{336}{625},\frac{840}{841},\frac{840}{1369},\frac{720}{1681},\frac{2520}{2809},\frac{1320}{3721},\frac{2016}{4225},\frac{3696}{4225},\frac{5280}{5329},\frac{2184}{7225},\frac{5544}{7225},\frac{6240}{7921},...$$

The denominators (the value of $c$, not $c^2$) seem to correspond directly to the "Ordered hypotenuses (with multiplicity) of primitive Pythagorean triangles" OEIS A020882 and the numerators to "Common differences in triples of squares in arithmetic progression, that are not a multiples of other triples in (A$198384$, A$198385$, A$198386$)" OEIS A198438.

With this information, I am unsure how to prove that these sequences will enumerate a full rational parameterization of my two initial expressions without missing any rational points, and how to generate a parameterization of these rational values.

Final Notes

Any hints, ideas, or references would be much appreciated!

Edit

Thanks to John Omielan and using my techniques above, I have determined that $\sqrt{1-x}$ and $\sqrt{1+x}$ are rational when $x=\frac{4t(t^2-1)}{(t^2+1)^2}$ for all rational values of $t$.

Answer 1

For $\sqrt{c^2 - a}$ and $\sqrt{c^2 + a}$ to be integers means for some integers $b$ and $d$ you have

$$\sqrt{c^2 - a} = b \implies c^2 - a = b^2 \tag{1}\label{eq1A}$$

$$\sqrt{c^2 + a} = d \implies c^2 + a = d^2 \tag{2}\label{eq2A}$$

\eqref{eq2A} minus \eqref{eq1A} gives

$$2a = d^2 - b^2 \implies a = \frac{d^2 - b^2}{2} \tag{3}\label{eq3A}$$

Adding these $2$ equations instead gives

$$2c^2 = b^2 + d^2 \tag{4}\label{eq4A}$$

There are a couple of good answers in Quora's Are there two squares that when added the sum is twice another square? that help to solve this. First, Justin Rising's answer explains

We start by noting that $a^2 + b^2 = 2c^2$ if and only if $\left(\frac{a}{c\sqrt{2}}\right)^2 + \left(\frac{b}{c\sqrt{2}}\right)^2 = 1$. This means that the point $(\frac{a}{c\sqrt{2}}, \frac{b}{c\sqrt{2}})$ lies on the unit circle. If we rotate it by $\frac{\pi}{4}$ radians, we get $(\frac{a}{2c} − \frac{b}{2c}, \frac{a}{2c} + \frac{b}{2c})$. Therefore, every solution to the the original equation corresponds to a rational point on the unit circle.

Next, Ben Packer's answer extends this to show that every rational point on the unit circle corresponds to a Pythagorean triple, i.e.,

$$x^2 + y^2 = z^2 \tag{5}\label{eq5A}$$

Then setting

$$\frac{b}{c} = -\frac{x}{z} + \frac{y}{z} \tag{6}\label{eq6A}$$

$$\frac{d}{c} = \frac{x}{z} + \frac{y}{z} \tag{7}\label{eq7A}$$

gives a solution to \eqref{eq4A}. Note this connection to Pythagorean triples helps to explain your observation of

The denominators (the value of $c$, not $c^2$) seem to correspond directly to the "Ordered hypotenuses (with multiplicity) of primitive Pythagorean triangles" OEIS A020882