Timeline for A question about integer triples

6 events

when toggle format	what		by	license	comment
Dec 24, 2019 at 7:26	answer	added	Tomita		timeline score: 1
Dec 17, 2019 at 20:25	answer	added	Will Jagy		timeline score: 0
Dec 16, 2019 at 18:53	comment	added	Vlad Matei		To parametrize the ellipse note that you have the obvious rational point $(0,2)$ and any other rational point on the ellipse will determine a line with this rational point of rational slope. Thus we can write all the solutions as $X=t(Y-2)$ for $t\in\mathbb{Q}$. This gives $Y=\cfrac{6t^2-2}{3t^2+1}$ and $X=\cfrac{4t}{3t^2+1}$.
Dec 16, 2019 at 18:44	comment	added	Vlad Matei		To finish off you can write these solutions in parametric form. Namely note that this intersection is an ellipse so it should have a nice form. To do it substitute $c=1-a-b$. This give $a^2+b^2+ab-a-b=0$ and the equation of the underlying ellipse is $3X^2+Y^2=4$ where $X=2a+b-1$ and $Y=3b-1$.
Dec 16, 2019 at 18:34	comment	added	Vlad Matei		So there are "obvious" solutions when both $xy+yz+zx=pq+qr+rs=0$ or $x+y+z=p+q+r=0$ . Otherwise note that you can rewrite the above as being $\sum (\frac{x}{x+y+z})^2=\sum (\frac{p}{p+q+r})^2$. Now you can think about this as looking for rational points on the intersection of the sphere $a^2+b^2+c^2=1$ with $a+b+c=1$.
Dec 16, 2019 at 15:45	history	asked	Clark Kimberling	CC BY-SA 4.0