I observe that $O(1,0)$ is such a rational pair. I now consider this as an ellipse in cartesian 2-space. If a point $A(x,y)$ on it is rational, then $y/x-1$ will be rational (unless x=1, but this case has already been addressed), so the line $AO$ has rational gradient. Conversely, if I take any line $y=mx-m$ through $O$ with $m\in\mathbb{Q}$, it meets the ellipse again at $(m^{2}+1)x^{2}+2m^{2}x+m^{2}-1=0$ where $x=\frac{-2m^{2}\pm \sqrt{4m^{4}-4(m^{2}+1)(m^{2}-1)}}{2(m^{2}+1)}=\frac{-m^{2}\pm 1}{m^{2}+1}$, where $y=mx-m=\frac{-2m^{3}-m \pm m}{m^{2}+1}$ which is clearly a rational point. By arguing both ways I have shown that all rational points on the given ellipse are of the form above, so the solution is \[(x,y)=\left(\frac{-m^{2}\pm 1}{m^{2}+1},\frac{-2m^{3}-m \pm m}{m^{2}+1}\right)\ \ \forall m \in \mathbb{Q}\]

Moonmathpi496 wrote: Nice solution! if they asked to find int solution.The problem would be 100 times easier...But actually i dont know how to find rational solution. How to do it?Using cartesian geometry?Or is there any general way? After all,how can i learn it? Thank u Moon

In fact, there exists an elementary approach making use of no analytic geometry. Let $ (x, y)$ be a solution to our equation. If some component is zero, we ascertain $ (x, y) = (\pm 1, 0)$. Of course that $ (\pm x, \pm y)$ are also solutions, so let's assume now $ x, y > 0$. Then write $ x = \frac{p}{q}$, $ y = \frac{r}{s}$ with $ p, q, r, s \in \mathbb{N}$ and $ (p, q)$, $ (r, s)$ coprime pairs. We rewrite the equation as $ p^2s^2 + 3r^2q^2 = q^2s^2$. Notice that $ q^2$ divides $ p^2s^2$ is implying $ q^2 | s^2$, and then $ s^2 | 3r^2q^2$ yields $ s^2 | 3q^2$. So $ 3q^2 = ms^2 = mnq^2$, $ \{m, n\} \subset \mathbb{N}$, hence either $ s^2 = 3q^2 \Rightarrow \sqrt{3} = \frac{s}{q}$, known to be false, or $ q = s$. It has just been inferred that the equation is reduced to $ p^2 + 3r^2 = q^2$. The methods from now on are varied, however I prefer the idea to factor as $ (p - r)(p + r) = (q - 2r)(q + 2r)$. By the Four-Number Theorem, there exists an unique quadruple of integers $ (a, b, c, d)$ such that $ a > 0$, $ b, d$ are of greatest common divisor $ 1$ and $ p - r = ab$, $ p + r = cd$, $ q - 2r = ad$ and $ q + 2r = bc$ hold. Thus $ p = \frac{cd + ab}{2}$, $ r = \frac{cd - ab}{2}$, $ q = \frac{bc + ad}{2}$ and $ r = \frac{bc - ad}{4}$. Equating the two formulas for $ r$ yields $ 2(cd - ab) = bc -ad$, or equivalently $ (2c + a)d = (2a + c)b$. Now since $ (b, d) = 1$, $ 2c + a = bv$ and $ 2a + c = dv$ for some integer $ v$. By solving the system, $ a = \frac{(2d - b)v}{3}$ and $ c = \frac{(2b - d)v}{3}$. Replacing in the previous relations finally leads us to $ p = \frac{v}{6} \cdot (4bd - b^2 - d^2)$, $ q = \frac{v}{6} \cdot (2b^2 + 2d^2 - 2bd)$ and $ r = \frac{v}{6} \cdot (b^2 - d^2)$. In the end, for convenience rename the parameters $ m : = b$, $ n : = d$ and replace them in the formulas for $ p, q, r$ and then $ x, y$ to ascertain the general form of the solution $ (x, y) = \left(\pm \frac{4mn - (m^2 + n^2)}{2(m^2 + n^2 - mn)} , \pm \frac{m^2 - n^2}{2(m^2 + n^2 - mn)} \right)$, for all $ (m, n) \in \mathbb{Z}^2 \backslash \{ 0, 0 \}$.

Let x=p/q, y=r/s （p,q,r,s∈N） and （p,q）,（r,s） are coprime pairs.By above PhilAndrew's solution, p^2＋3r^2=q^2 holds.From now, my method is different. （q＋p）（q－p）=3r^2 g=G.C.D（q＋p,q－p） is ≦2. Because if g is ≧3, g | 2p and ∃g' （≧2）| g s.t. g' | p.So g' | q holds, this is contradiction.Then g is ≦2.Apparently q＋p, q－p are same parity. Case1）q＋p, q－p are odd Case1-1）3 | q＋p We can write q＋p=3m^2, q－p=n^2 with（m,n） are coprime odd pairs, and 3m^2＞n^2.Then p=（3m^2－n^2）/2, q=（3m^2＋n^2）/2, r=mn.So x=（3m^2－n^2）/（3m^2＋n^2）, y=2mn/（3m^2＋n^2） with （m,n） are coprime odd pairs, and 3m^2＞n^2.These satisfy the equation. Case1-2）3 | q－p We can write q＋p=n^2,q－p=3m^2 with （m,n） are coprime odd pairs, and 3m^2＜n^2.Then p=（n^2－3m^2）/2, q=（3m^2＋n^2）/2, r=mn.So x=（n^2－3m^2）/（3m^2＋n^2）, y=2mn/（3m^2＋n^2） with （m,n） are coprime odd pairs, and 3m^2＜n^2.These satisfy the equation. By Case1-1） , Case1-2）, We write the solution, x=|（3m^2－n^2）/（3m^2＋n^2）|, y=2mn/（3m^2＋n^2） with （m,n） are coprime odd pairs---α. Case2）q＋p, q－p are even Clearly q＋p and q－p are distinct in mod 4. Case2-1）q＋p≡0, q－p≡2 mod 4 Case2-1-1）3 | q＋p We can write q＋p=3・2^a・m^2, q－p=2n^2 with a≧2,（m,n） are coprime odd pairs, and 3・2^（a－1）・m^2＞n^2.Then p=3・2^（a－1）・m^2－n^2, q=3・2^（a－1）・m^2＋n^2, r=2^｛（a＋1）/2｝・mn.So x=（3・2^（a－1）・m^2－n^2）/（3・2^（a－1）・m^2＋n^2））, y=（2^｛（a＋1）/2｝・mn）/（3・2^（a－1）・m^2＋n^2） with a≧2, （m,n） are coprime odd pairs, and 3・2^（a－1）・m^2＞n^2. Case2-1-2）3 | q－p Similarly to above Case, x=｛2^（a－1）・n^2－3m^2｝/｛2^（a－1）・n^2＋3m^2｝, y=（2^｛（a＋1）/2｝・mn）/｛2^（a－1）・n^2＋3m^2｝ with a≧2, （m,n） are coprime odd pairs, and 3m^2＜2^（a－1）・n^2. Case2-2）q＋p≡2, q－p≡0 mod 4 Case2-2-1）3 | q＋p Similarly to above Case, x=（3m^2－2^（a－1）・n^2）/｛3m^2＋2^（a－1）・n^2｝, y=（2^｛（a＋1）/2｝・mn）/｛3m^2＋2^（a－1）・n^2｝ with a≧2, （m,n） are coprime odd pairs, and 3m^2＞2^（a－1）・n^2. By Case2-1-2）, Case2－2－1）, we write the solution, x=|（3m^2－2^（a－1）・n^2）/｛3m^2＋2^（a－1）・n^2｝|, y=（2^｛（a＋1）/2｝・mn）/｛3m^2＋2^（a－1）・n^2｝ with a≧2, （m,n） are coprime odd pairs---β. Case2-2-2）3 | q－p Similarly to above Case, x=｛n^2－3・2^（a－1）・m^2｝/｛n^2＋3・2^（a－1）・m^2｝, y=（2^｛（a＋1）/2｝・mn）/｛n^2＋3・2^（a－1）・m^2｝ with a≧2, （m,n） are coprime odd pairs, and 3・2^（a－1）・m^2＜n^2. By Case2-1-1）, Case2-2-2）, We write the solution, x=|｛3・2^（a－1）・m^2－n^2｝/｛3・2^（a－1）・m^2＋n^2｝|, y=（2^｛（a＋1）/2｝・mn）/｛3・2^（a－1）・m^2＋n^2｝ with a≧2, （m,n） are coprime odd pairs---γ. Therefore the solution are α,β,γ form.

Hey, IIthigore. y=mx－m meets the ellipse at （3m^2＋1）x^2－6m^2・x＋（3m^2－1）=0 where x=（3m^2±1）/（3m^2＋1）, y=（－m±m）/（3m^2＋1）? So the solution are x=（3m^2－1）/（3m^2＋1）, y=（－2m）/（3m^2＋1） with m＜－1/3^1/2.