The first we has a result : Let $ \gcd(m,n) = 1$ then $ \gcd(m^3 + n^3,m^2 + n^2)\in \{1,2\}$ Proof Let $ d = \gcd(m^2 + n^2,m^3 + n^3)$ Imply that $ p|2x^6 \Rightarrow d|2$ so we has result. Now consider our solution. Let $ x = \frac {m}{p},y = \frac {n}{q}$ $ p = \frac {m^3 + n^3}{m^2 + n^2}$ Let $ d = \gcd(m,n)$ then $ m = da,n = db$ where $ \gcd(a,b) = 1$ So $ p = d\frac {a^3 + b^3}{a^2 + c^2}$ Case 1 $ \gcd(a^3 + b^3,a^2 + b^2) = 1$ so $ a^2 + b^2|d$ Imply that $ d = k(a^2 + b^2)$ $ p = k(b^3 + a^3)$ so $ x = \frac {da}{p} = \frac {(a^2 + b^2)a}{a^3 + b^3}$ $ y = \frac {db}{p} = \frac {(a^2 + b^2)b}{a^3 + b^3}$ Case 2 $ \gcd(a^3 + a^3,a^2 + b^2) = 2$ it mean that $ a\equiv b\equiv 1(\mod 2)$ Then $ a^2 + b^2|2d$ imply that $ d = \frac {k(a^2 + b^2}{2}$ $ p = \frac {k(a^3 + b^3)}{2}$ $ x = \frac {a(a^2 + b^2)}{a^3 + b^3}$ $ y = \frac {b(b^2 + a^2)}{a^3 + b^3}$ Result : $ x = \frac {a(a^2 + b^2)}{a^3 + b^3}$ $ y = \frac {b(a^2 + b^2)}{a^3 + b^3}$

Eh? So you say that $ a=100,b=101$ gives solution $ x=2020100/2030301$, $ y=101$? Because that really ain't no solution. Just looking at the powers appearin in your solution says that you have a problem for big $ a,b$...

I am sorry my solution is true but I have a mistake when I write it.

You are right. My solution is wrong (and has been deleted).

I am sorry that I am asking almost a year after, but which are the solutions of this equation, because it appeared in a Bulgarian Olympiad a year ago and I don't know how to solve it.

Assume that $ x = \frac {a}{b}, y = \frac {c}{d}$ is a solution, and (a,b)=(c,d)=1. Let $ k = (b,d), t = d/k, s = b/k$. If x=0 or y=0: it's easy to see that the possible solutions with zeros are $ (0,0),(0,1),(1,0)$. Now assume that a,b,c,d are non-zero integers. $ x^3 + y^3 = x^2 + y^2$ is equivalent to $ (at)^3 + (cs)^3 = kts((at)^2 + (cs)^2)$, which implies that $ (at)^2 + (cs)^2 | (at)^3 + (cs)^3$. But, as TTsphn showed, it implies that $ (at)^2 + (cs)^2 = 1$ or $ (at)^2 + (cs)^2 = 2$. First case: $ (at)^2 + (cs)^2 = 1 \implies$, WLOG: $ |cs| = 0$, a contradiction. Second case: $ (at)^2 + (cs)^2 = 2 \implies |a| = |t| = |c| = |s| = 1$, or: $ |a| = |c| = 1, |b| = |d|$. We can assume WLOG a,c>0, and then: a=c=1. Because $ x^2 + y^2 > 0$, we must have b=d, which gives us: $ x^3 + y^3 = \frac {2}{b^3} = x^2 + y^2 = \frac {2}{b^2}$, or: b=1. All the solutions are: $ (0,0),(0,1),(1,0),(1,1)$

put $ \frac{x}{y}=a \in Q$ (the case $ y=0$ is trivial ) then we get : $ (ay)^3+y^3=(ay)^2+y^2 \implies y=\frac{a^2+1}{a^3+1} \implies x=\frac{a^3+a}{a^3+1}$ so we have either $ (x,y)=(\frac{a^3+a}{a^3+1},\frac{a^2+1}{a^3+1})$ where $ a \in Q$ and $ a \neq -1$. and by considering the case $ a=-1$ we get the solution $ (x,y)=(0,0)$ so finally we have $ (x,y)=(\frac{a^3+a}{a^3+1},\frac{a^2+1}{a^3+1}),(0,0),(1,0)$ where $ a \in Q$ and $ a \neq -1$ and $ (1,0)$ comes from the case $ y=0$.

@ Shoki : yeah this is the General form of x,y but you should find all $a$s then...

@ Shoki : yeah this is the General form of x,y but you should find all $a$s then... @TTsphn : haha, so you will get that a^3(b-1) = b^3(1-a), then what? @bambaman : i don't know how did you use bambaman wrote: as TTsphn showed, it implies that $ (at)^2 + (cs)^2 = 1$ or $ (at)^2 + (cs)^2 = 2$. cuz i don't think it's true, or from where it came from. ...

There has to be a rational parametrisation because it is a smooth projective geometrically connected curve of genus $0$ with a rational point, hence isomorphic to $\mathbf{P}^1_\mathbf{Q}$.