find all rational numbers $x,y$ such that we have : $$x^3+y^3=x^2+y^2$$
my attempt :
first i wrote $x$ as $\frac{\alpha}{\beta}$ and $y$ as $\frac{s}{t}$ $(\gcd(\alpha,\beta)=\gcd(s,t)=1)$ and tried to solve it with it's new face but i couldn't .
then i wrote $x$ as $ay$ ($x,y\in\mathbb{Q}$ so $\frac{x}{y}=a\in\mathbb{Q}$ and $x=ay$) then $$(ay)^3+y^3=(ay)^2+y^2$$$$y(a^3+1)=a^2+1$$$$x=ay=\frac{a^3+a}{a^3+1}$$
On
This cubic has an isolated point at $(0,0)$; it is singular there. We can confirm this by checking that $f(x,y)=x^3+y^3-x^2-y^2$ is $0$ at that point and at its partial derivatives, $f(0,0)=f_x(0,0)=f_y(0,0)=0$. To be clear for the $x$ partial derivative we have: $f_x(x,y)=3x^2-2x=0$ at that point.
Now a trick we can use is: a line intersects a cubic at 3 rational points if it intersects at 2 rational points. Since $(0,0)$ is a singular point, specifically a double point, that means the line $y=tx$ for some rational slope $t$ will intersect at exactly one other point. This means we've parametrized nearly the entire set of rational points!
We can now plug in and solve for $x,y$:
$$x^2+(tx)^2 = x^3 + (tx)^3$$ Since we're interested in the 3rd point, not the singular point $(0,0)$ we can divide by $x \ne 0$,
$$1+t^2=x(1+t^3)$$ $$x=\frac{1+t^2}{1+t^3}$$ Getting $y$ is immediate since $y=tx$. And here is our parametrization of rational points:
$$\left(\frac{1+t^2}{1+t^3},t\frac{1+t^2}{1+t^3}\right)$$
Keep in mind there is one point we can't get from parameterization unless we allow the slope to be "infinite," since a vertical line will intersect at the point $(0,1)$.
By inspection there are trivial solutions $(0,0), (0,1), (1,0)$; moreover, the equation is cubic, hinting that there are infinitely many rational solutions. Some examples: (10/7, -5/7); (15/13, -5/13); (39/19, -26/19); (100/37, -75/37); (85/49, -51/49); (205/61, -164/61); (65/62, -13/62); (68/63, -17/63); (366/91, -305/91).
Apart from in the trivial solutions above, $x$ and $y$ have opposite signs; wlog $y<0<x$. I'd rather work in integers than in rationals, so I consider numerators and denominators separately. Let $d$ be $x$'s and $y$'s lowest common denominator, and $c$ the highest common factor of the resulting numerators. Then $x=ac/d, y=-bc/d$ for some $a$ and $b$.
\begin{align*} \frac{a^3c^3}{d^3}-\frac{b^3c^3}{d^3} &= \frac{a^2c^2}{d^2}+\frac{b^2c^2}{d^2}\\ \implies (a^3-b^3)c^3 &= (a^2+b^2)c^2d\\ \implies (a^3-b^3)c &= (a^2+b^2)d\\ \implies \frac{a^3-b^3}{a^2+b^2} &= \frac{d}{c} \end{align*}
meaning that coprime $a>0$ and $b>0$ may be chosen at will, determining $c$ and $d$ (given that $c/d$ is in its lowest terms).