Finding ALL rational solutions to a diophantine equation I came accross a problem in an excerpt of the book "1001 problems in classical number theory" that asks to determine all rational solutions to the equation :
$$ x^3 + y^3 = x^2 + y^2$$ 
Apart from the trivial solutions $(1,0)$ $(1,1)$, $(0,1)$ $(0,0)$, I couldn't find a way to determine all rational solutions. 
Here are some other solutions found by some algorithm >> https://benvitalenum3ers.wordpress.com/2015/07/01/rational-number-solutions-to-x3-y3-x2-y2/
Any ideas are welcome.
Thanks
 A: I get that all solutions are the intersection of the curve (over the reals) with lines through $(1,1)$ with slope a negative square, that is
$$ p^2 x + q^2 y = p^2 + q^2  $$
with intgers $p,q > 0$ and, might as well, $\gcd(p,q) = 1.$
ADDED: I went through it more carefully, for each positive pair we get two intersections. A little cleaner to allow integers $\gcd(p,q) = 1$ and $p + q \neq 0,$ not necessarily both positive. Then
$$ x = \frac{ p^3  + pq^2\; }{p^3 + q^3}= \frac{ \left(p^2  + q^2\right)p \; }{p^3 + q^3} \; \; , $$
$$ y = \frac{p^2 q + q^3 \;}{p^3 + q^3} \; \; = \frac{ \left(p^2  + q^2\right)q \; }{p^3 + q^3} \; \; .  $$
For that matter, it turns out that we can take any direction $(p,q)$ with integers $p,q$ and $p+q > 0$ ( so that we will also have $p^3 + q^3 > 0$) and simply adjust the length by multiplying by the correct scalar quantity, giving
$$ (x,y) = \; \; \left(  \frac{ \; p^2  + q^2 \; }{\; p^3 + q^3 \;} \right) \; \; \; (p,q) $$
Let me edit in the nine examples at LINK
For example, with $(p,q) = (3,-1),$ we get $\frac{ \; p^2  + q^2 \; }{\; p^3 + q^3 \;} = \frac{9+1}{27 - 1} = \frac{10}{26} = \frac{5}{13},$ so we get $(x,y) =  \left(\frac{5}{13}\right) (3,-1) = \left(\frac{15}{13}, \; \; -\frac{5}{13} \right)$
$$ (3,-1) \mapsto \left( \frac{15}{13}, \; -  \frac{5}{13} \right)   $$
$$ (2,-1) \mapsto \left( \frac{10}{7}, \; -  \frac{5}{7} \right)   $$
$$ (5,-3) \mapsto \left( \frac{85}{49}, \; -  \frac{51}{49} \right)   $$
$$ (3,-2) \mapsto \left( \frac{39}{19}, \; -  \frac{26}{19} \right)   $$
$$ (7,-5) \mapsto \left( \frac{259}{109}, \; -  \frac{185}{109} \right)   $$
$$ (4,-3) \mapsto \left( \frac{100}{37}, \; -  \frac{75}{37} \right)   $$
$$ (9,-7) \mapsto \left( \frac{585}{193}, \; -  \frac{455}{193} \right)   $$
$$ (5,-4) \mapsto \left( \frac{205}{61}, \; -  \frac{164}{61} \right)   $$
$$ (11,-9) \mapsto \left( \frac{1111}{301}, \; -  \frac{909}{301} \right) $$
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Indeed, all we need to do is take consecutive exponents, say
$$ \color{red}{x^7 + y^7 = x^6 + y^6.} $$
We can take any direction $(u,v)$ with integers $u,v$ and $u+v > 0$ and  multiply by the correct scalar quantity, giving
$$ (x,y) = \; \; \left(  \frac{ \; u^6  + v^6 \; }{\; u^7 + v^7 \;} \right) \; \; \; (u,v) $$
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The precise form of the numerator and denominator are not that important, as long as they are homogeneous. I would get an analogous outcome for
$$\color{red}{ x^4 + x^2 y^2 + y^4 = x^3 + x^2 y + x y^2 + y^3}, $$ all rational points are
$$ (x,y) = \; \; \left(  \frac{ \; u^3 + u^2 v + u v^2 + v^3 \; }{\;   u^4 + u^2 v^2  + v^4   \;} \right) \; \; \; (u,v) $$

