Find all rational points on the elliptic curve $$ y^2=x^3-\frac{13}{3} x+\frac{70}{27}$$
The problem is essentially equivalent to the following problem;
Find al pairs $(x,y)$ of rational numbers such that $xy^2=x^2+2x-3$.
This is from the Korea Junior mathematical Olympiad held in 2022. The problem and its proposed unofficial solutions could be found in KJMO 2022 P8
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This problem was quite difficult as a math olympiad problem for middle school students. According to an unofficial source, not a single participant received even partial credit for this problem.
This problem could be transformed into one finding all rational points on an elliptic curve. Note that $x$ never be zero. Hence if we substitute $y$ for $\frac{y}{x}$ (with abusing use of variable) and multiply the equation by $x$, we have an equivalent equation which in turn, by substituting $x$ for $x- \frac{2}{3}$ transformed into the equation $$ y^2=x^3-\frac{13}{3} x+\frac{70}{27}$$
Sometimes math olympiad problems can be solved easily with modern mathematics. My point is that this problem may be one of the case. Therefore, let us examine whether the elliptic curve theory can solve this problem manually.
This is possible using the technique from the first three chapters of Silverman & Tate for proving the Mordell's theorem where the elliptic curve has $2$-torsion. Although it's unnecessarily complicated for the current purpose, it might be useful for other problems. I will sketch the main definitions and results without proofs. All the cited results in the following are referring to the ones from S&T.
It's better just to consider the curve $y^2=f(x)=x^3+2x^2-3x$.
By the Nagell–Lutz Theorem (Theorem 2.5), rational torsion points on $y^2=x^3+ax^2+bx+c$ where $a,b,c\in\mathbb Z$ must have integer coordinates, and further $y=0$ or $y|D$ (the discriminant of $f(x)$). Therefore the problem is reduced to finding integer solutions of finitely many cubic equations. In our case, by the rational root theorem, $x$ must be a divisor of $y^2$, hence $D^2=(2^4\cdot 3^2)^2$. So we may substitute $x$ with all divisors of $D^2$ to see whether $f(x)$ is a complete square. Details are omitted.
Now it remains to show that the given curve has rank $0$. This is the hard part. There might be easier way, but my understanding of elliptic curves is quite dusty now, so I will just try to follow S&T.
Consider the curve $$C:y^3=x^3+ax^2+bx, \Gamma:=C(\mathbb Q)$$ where $a,b\in\mathbb Z$, and the induced one $$\overline{C}: y^3=x^3+(-2a)x^2+(a^2-4b)x, \overline{\Gamma}:=\overline{C}(\mathbb Q)$$
By Proposition 3.7, we can construct a pair of homomorphisms: $$\phi: \Gamma\rightarrow \overline\Gamma, \psi:\overline\Gamma\rightarrow \Gamma$$ such that $$\psi\circ \phi (P)=2P, \phi\circ \psi (Q)=2Q$$
Combined with Mordell's theorem, we can show that (Section 3.6) $$2^r\cdot\#\Gamma[2]=[\Gamma:2\Gamma]=[\Gamma:\psi(\overline\Gamma)][\overline\Gamma:\phi(\Gamma)]$$ Where the $2$-torsion group $\Gamma[2]$ consists of $\mathcal O$ and all points with $y=0$.
We won't need the exact definition of $\phi,\psi$ (which is explicitly given in S&T) but we do need to know more details of the following homomorphism to help analyze $[\Gamma:\psi(\overline{\Gamma})]$ and $[\overline\Gamma:\phi(\Gamma)]$ through Proposition 3.8, $$\alpha:\Gamma\rightarrow\mathbb Q^*/(\mathbb Q^*)^2$$
such that $$\begin{cases} \alpha(\mathcal O)=1 & \text{ mod } (\mathbb Q^*)^2 \\ \alpha(0, 0)=b & \text{ mod } (\mathbb Q^*)^2 \\ \alpha(x\not=0, y)= x & \text{ mod } (\mathbb Q^*)^2 \end{cases}$$
And it was shown,
In our particular case, $a=2, b=-3$ and $(-2a)=-4, (a^2-4b)=16$.
So $\Gamma[2] = \{\mathcal O, (0,0), (1, 0), (-3, 0)\}$.
By Proposition 3.8 (d), as $b=-3$ has a unique prime factor, $[\Gamma:\psi(\overline\Gamma)]\le 2^2=4$.
By 3.8 (c), the image of $\overline{\alpha}:\overline\Gamma\rightarrow \mathbb Q^*/(\mathbb Q^*)^2$ is contained in $\{\pm 2(\mathbb Q^*)^2\}$. Note that for $$\overline C: y^3=x^3-4x^2+16x=x(x^2-4x+16)=x((x-2)^2+12)$$ $x$ can never be negative (and $a^2-4b=16>0$ as well), therefore the image of $\overline\alpha$ doesn't contain $-1$ mod $(\mathbb Q^*)^2$, hence $[\overline\Gamma: \phi(\Gamma)]=|\overline\alpha(\overline\Gamma)|=1$. Now it follows that $r=0$.