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.
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1$\begingroup$ To clarify, I am not asking for conditions on the fraction $r/s$ to be a Pythagorean fraction, I am asking for conditions on a fraction that make it a ratio of two Pythagorean fractions, i.e., of the form $\frac{a/b}{d/e}$ where $a^2 + b^2$ and $d^2 + e^2$ are both perfect squares. $\endgroup$– Brian LinsMay 15, 2013 at 0:18
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$\begingroup$ I updated my response. $\endgroup$– GH from MOMay 15, 2013 at 1:20
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2 Answers
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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.
answered May 15, 2013 at 0:05
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1$\begingroup$ Am I misreading the question? Do you want to know if $4/9$ is a RATIO of two Pythogorean fractions, as opposed to being a Pythagorean fraction by itself? The wording of your question means an expression for $4/9$ as $\frac{p_1q_1/(p_1^2-q_1^2)}{p_2q_2/(p_2^2-q_2^2)}$ is what is desired (using Barry Cipra's suggestion). $\endgroup$ May 15, 2013 at 0:26
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2$\begingroup$ This elliptic curve is 120a1 in Cremona's tables, it has rank 0 and its Mordell-Weil group is isomorphic to $\mathbf{Z}/4\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z}$. So it seems that the only rational points on this intersection of 2 quadrics are those with $x=0$ or $y=0$. $\endgroup$ May 15, 2013 at 9:30
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1$\begingroup$ Sorry, this is 120a2 and not 120a1. $\endgroup$ May 15, 2013 at 9:30
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1$\begingroup$ The conclusion is that 4/9 is not the ratio of any two Pythagorean fractions. $\endgroup$ May 15, 2013 at 9:32
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1$\begingroup$ For those interested in putting an intersection of two quadrics into cubic and then Weierstrass form, you can have a look at the first 2 pages of the following note I wrote perso.ens-lyon.fr/francois.brunault/recherche/biquadratique.pdf The equations are not precisely the same, but you can easily adapt and find the right change change of variables. $\endgroup$ May 15, 2013 at 9:42
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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.
answered May 14, 2013 at 23:38