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This question already has an answer here:

How to find the number of rational points on the circumference of ellipse $$ \frac{x^2}{9}+\frac{y^2}{4}=1 \,$$

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marked as duplicate by Eevee Trainer, StackTD, Arthur, Dietrich Burde, Community Apr 4 at 7:11

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  • $\begingroup$ What have you tried? $\endgroup$ – Arthur Apr 2 at 8:53
  • $\begingroup$ i tried making perfect square. but when for rational if i use y as p/q , i have feeling it becomes infinity? $\endgroup$ – maveric Apr 2 at 8:54
  • $\begingroup$ Hint: rational points on the circumference of ellipse $ \iff $ rational points on the boundary of the unit disk for your ellipse. $\endgroup$ – Bach Apr 2 at 9:01
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Use the fact there are infinitely many rational points on the unit circle (Are the rationals on a unit circle dense?), moreover of a classical form :

$$\left(\dfrac{m^2-n^2}{m^2+n^2} \ ; \ \dfrac{2mn}{m^2+n^2}\right)\tag{1}$$

($m,n$ any integers, not both zero).(https://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple)

Then use the following matrix with rational coordinates that maps the unit circle onto the ellipse to obtain :

$$\begin{pmatrix}3&0\\0&2\end{pmatrix}\begin{pmatrix}\tfrac{m^2-n^2}{m^2+n^2}\\ \tfrac{2mn}{m^2+n^2}\end{pmatrix}=\begin{pmatrix}\tfrac{3(m^2-n^2)}{m^2+n^2}\\ \tfrac{4mn}{m^2+n^2}\end{pmatrix}\tag{2} $$ Relationship (2) constitute a general expression for all the points with rational coordinates on the ellipse because the correspondence with points with rational coordinates on the circle is bijective through the transformation with matrix above.

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There are infinitely many. Let

$$C:\frac{x^2}9+\frac{y^2}4=1$$

We have $(3,0)\in C(\mathbb{Q})$. Now consider the straight line

$$L_t:y=t(x-3)$$

That passes through $(3,0)$ and has a rational slope $t$, working by hand $L_t\cap C$ you get that the intersection is a rational points, this is given by the fact that you have a system of equations that simplifies to a quadratic on $x$ which we know will have one rational solution ($x=3$) and therefore the other solution must be rational. Then, it is easy to see that we have a bijection between $C(\mathbb{Q})-\lbrace (3,0)\rbrace$ and $\mathbb{Q}$ given by

$$t\in\mathbb{Q}\to the\;\; only\;\; point\;\;on\;\; L_t\cap C\;\; besides\;\;(3,0)$$

So you have a rational point for every rational number, i.e. infinitely many rational points. Moreover, if you do this process step by step you can parametrize every rational point on the curve except for $(0,3)$.

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