Analyzing a specific function

Question

I was trying to solve a proving question and I finally arrived at this step where I am stuck. The statement is proved if there exists infinitely many ordered pairs $x_a$ and $x_b$ such that: $$(x_a-x_b)^2+(\sqrt{1-x_a^2}-\sqrt{1-x_b^2})^2=\frac{p^2}{q^2}$$ Where $p$ and $q$ are natural numbers and $x_a,x_b\in[-1,1]$. How would you go on to prove such a fact?

Answer 1

(I answered approximately the same as the first paragraph to a similar question you asked yesterday)

Clearly the LHS attains an infinite amount of real values, as $x_a$ and $x_b$ can attain any real value from −1 to 1. Because the LHS is continuous and we know that the rationals are dense in the reals, we can say that the LHS attains infinitely many rational values. Because there are obviously infinitely many perfect squares we then know that the LHS attains infinitely many square rationals(or whatever the term is). So, we have that your statement is true.

This may feel somewhat unsatisfying, so here's a different approach to your original question you posted in the comments about infinite points on the unit circle.

First, let one of the points be $(1, 0)$ and let the other point be $(a, b)$. We can say this because if this is not the case, we can "rotate" the unit circle along with the points until one of them ends up at $(1, 0)$. This will preserve the distance. Anyway, because $(a, b)$ lies on the unit circle, we have that: $$a^2+b^2=1$$ $$b^2=1-a^2$$ $$b=\sqrt{1-a^2}$$ Here, we take the positive square root, so we assume the point not at $(1, 0)$ is above the x-axis. We can assume this because if it is not the case, we just reflect this point over the x-axis(it will still be on the unit circle). This will preserve the distance between that point and $(1, 0)$. So, the distance between the two points is: $$\sqrt{(1-a)^2 + (0 - \sqrt{1-a^2})^2}=$$ $$\sqrt{a^2-2a+1 + 1 - a^2}=$$ $$\sqrt{2-2a}$$

Now, we can let the distance be $\frac{p}{q}$. So, we have that: $$\sqrt{2-2a}=\frac{p}{q}$$ $$2-2a=\frac{p^2}{q^2}$$ $$1-a=\frac{p^2}{2q^2}$$ $$a = 1 - \frac{p^2}{2q^2}$$

Then, we have that $b=\sqrt{1-a^2}$, so $$b=\sqrt{1-(1-\frac{p^2}{2q^2})^2}$$ $$b=\sqrt{1-(\frac{p^4}{4q^4} - \frac{p^2}{q^2} + 1)}$$ $$b=\sqrt{\frac{p^2}{q^2} - \frac{p^4}{4q^4}}$$ $$b=\frac{p}{q} \sqrt{1-\frac{p^2}{4q^2}}$$

Now, we know that for any rational distance $\frac{p}{q} \leq 2$ there is a pair of points $(1, 0)$ and $(1 - \frac{p^2}{2q^2}, \frac{p}{q} \sqrt{1-\frac{p^2}{4q^2}})$ such that the distance between them is $\frac{p}{q}$. Because there are infinitely many rationals, there are infinitely many points on the unit circle such that the distance between any two of them is a rational number, and hence your statement is true.

Note: By un-rotating(remember we rotated one of the points to $(1, 0)$ earlier) our pair of points we found around the unit circle, we can get infinitely many points for a single rational distance. This should make some sense.