Possible ratios of Pythagorean fractions A Pythagorean fraction is a number of the form $a/b$ or $b/a$ where $a$ and $b$ are the legs of a Pythagorean triple. Are there simple necessary and/or sufficient conditions for determining whether a rational number can be expressed as a ratio of two Pythagorean fractions? As an example, I would be interested to know if $4/9$ can be expressed as a ratio of two Pythagorean fractions. A brute force search of all primitive Pythagorean triples $(a,b,c)$ with $c < 6000$ found that $4/9$ could not be written as a ratio of two Pythagorean fractions when one of the fractions has a corresponding hypotenuse less than 6000. 
 A: This is a sketch how to decide the question for $\frac{4}{9}$.
The question is if there are positive integers $a,b,d,e$ such that $\frac{4}{9}=\frac{a/b}{d/e}$ with $a^2+b^2$ and $d^2+e^2$ squares. Denoting $p:=\frac{9a}{b}=\frac{4d}{e}$, the question is if there is $p\in\mathbb{Q}^\times$ such that both $p^2+4^2$ and $p^2+9^2$ are rational squares. That is, the question is if the quadrics $x^2+4y^2=t^2$ and $x^2+9y^2=z^2$ over $\mathbb{Q}$ intersect in a point with $x,y\neq 0$. The intersection of the two quadrics is isomorphic to the elliptic curve 
$$ Y^2=8(X-1)(X+1)(9X-1) $$
according to a 1997 preprint by R.G.E Pinch: Square values of quadratic polynomials (which used to be here but is no longer available, unfortunately). If this elliptic curve has finitely many rational points (something that I cannot check at the moment for lack of time) then finding them explicitly will list all points $(x,y,t,z)$ lying on both quadrics, so the question if there is a point with $x,y\neq 0$ can be settled. Otherwise there surely will be a point with $x,y\neq 0$.
Added. It seems that François Brunault filled in the details (see his comments to this post), and $\frac{4}{9}$ is indeed not a ratio of two Pythagorean fractions.
A: Starting from the standard parameterization $(2pq,p^2-q^2,p^2+q^2)$ for primitive Pythagorean triples, we get $a/b = 2pq/(p^2-q^2)$ implies $(ap-bq)^2 = (a^2+b^2)q^2$ implies $(a^2+b^2)$ is a square.
Added later: Apologies, I didn't read the question carefully enough.  I thought I was giving an answer, but at best I was making a comment.
