Suppose $\frac{p}{q}$ and and $\frac{r}{s}$ are rationals in lowest terms (so $\gcd (p,q) = \gcd(r,s) = 1$) and $\frac{p^2}{q^2} + \frac{r^2}{s^2} = 1$; i.e. $p^2s^2+r^2q^2=q^2s^2$. Then exactly one of $p$ and $r$ is even, and $q$ and $s$ (and one of $p,r$) are odd.
Mainly, how do we find a contradiction when both numerators ($p$ and $r$) are odd and both denominators ($q$ and $s$) are even?
This section deals with all the other cases besides the one described in the previous sentence.
Suppose $p,r$ are both even. Then $qs$ must be even based on $(ps)^2+(rq)^2=(qs)^2$. But then $\gcd(p,q) \geq 2$ or $\gcd(r,s) \geq 2$. Therefore, $p$ and $r$ cannot both be even.
Suppose both $p$ and $r$ are odd. If both $q$ and $s$ are odd, then the left side of $(ps)^2+(rq)^2=(qs)^2$ will be even, and the right side will be odd. If exactly one of $q,s$ is even, then the left side of $(ps)^2+(rq)^2=(qs)^2$ will be odd, and the right side will be even.
The last case is when exactly one of $p,r$ is even. Since these fractions are in lowest terms, the denominator dividing the even numerator (dividing $p$ or $r$) must be odd. Suppose the other denominator (dividing the odd numerator) is even, so one of our two rationals $\frac{p}{q}$ and $\frac{r}{s}$ is $\frac{even}{odd}$, and the other is $\frac{odd}{even}$. But then the left side of $(ps)^2+(rq)^2=(qs)^2$ will be odd (it has an $even \cdot even$ term summed with an $odd \cdot odd$ term), and the right side will be even. So in this case, both $q$ and $s$ must be odd, as desired.
This question relates to a proof in D. R. Woodall, "Distances realized by sets covering the plane" (https://www.sciencedirect.com/science/article/pii/0097316573900204). This proposition helps prove that you can color $\mathbb{Q}^2$ with two colors such that no two points distance $1$ apart have the same color (the "chromatic number of the rational plane" is $2$). The chromatic number of $\mathbb{R}^2$, on the other hand, is $5,6$, or $7$.