<s>[INST] Find all $x,y$ such that $x^3+5y^3=(a^3+5b^3)^3$. Let $a,b$ be coprime integers. I am trying to find all integers $x,y$ such that:
$$x^3+5y^3=(a^3+5b^3)^3$$
What I have tried:
$$5y^3=(a^3+5b^3-x)[(a^3+5b^3)^2+(a^2+5b^3)x+x^2         ]$$
There are 2 Cases depending which factor of $5y^3$ is divisible by $5$:
Case 1: $5$ divides the first factor:
$$a^3+5b^3-x=5y_0^{3-n}y_1^3$$
$$(a^3+5b^3)^2+(a^2+5b^3)x+x^2 =y_0^ny_2^3$$
where $$y=y_0y_1y_2$$ $$(y_1,y_2)=1$$ $$n=\{0,1,2,3\}$$
It follows $$x=a^3+5b^3-5y_0^{3-n}y_1^3$$
We substitute $x$
$$(a^3+5b^3)^2+(a^2+5b^3)(a^3+5b^3-5y_0^{3-n}y_1^3)+(a^3+5b^3-5y_0^{3-n}y_1^3)^2 =y_0^ny_2^3$$
For simplicity, we write $u=a^3+5b^3$:
$$u^2+u(u-5y_0^{3-n}y_1^3)+(u-5y_0^{3-n}y_1^3)^2 =y_0^ny_2^3$$
$$u^2+u^2-5y_0^{3-n}y_1^3u+u^2-10uy_0^{3-n}y_1^3+25y_0^{2(3-n)}y_1^6=y_0^ny_2^3$$
$$3u^2-15uy_0^{3-n}y_1^3+25y_0^{2(3-n)}y_1^6=y_0^ny_2^3$$
What do I do at this point? Any input will be appreciated. Thanks.
 [/INST] The only solutions are the ones with $y=0$.
This is one of those problems where generalizing makes it easier, so I'm going to substitute the RHS with the cube of a general integer
$$x^3+5y^3=z^3$$
If $z=0$ you get the $(0,0,0)$ solution and otherwise $\frac x y =-5^{1/3}$ which is impossible. Now I'm going to let $x_1=x/z$ and $y_1=y/z$ so
$$x_1^3+5y_1^3=1$$
Now we just need to prove that $(1,0)$ is the only rational point on this curve. Let $t=x-1$ and $u=k(x+1)$. You can think of this as tracing a line of slope $k$ through the rational point we already know.
$$(t+1)^3+5(kt)^3=1$$
$$k^3=-\frac{t^2+3t+3}{5t^2}$$
Now let $t_1=1/t$ (note that from now on we're assuming $y \neq 0$)
$$3t_1^2+3t_1+1=-5k^3$$
$$3(2t_1+1)^2=-20k^3-1$$
Now to transform this into an elliptic curve, and abusing the notation by having a variable collision because that's what the software wants $x=60k$ and $y=180(2t_1+1)$ then
$$y^2=x^3-10800$$
Now we can get sage to tell us that this has no solutions:
sage: E = EllipticCurve([0,0,0,0,-10800])          
sage: E                                                                         
Elliptic Curve defined by y^2 = x^3 - 10800 over Rational Field                                 
sage: E.rank()                                                                  
0
sage: E.torsion_points()                                                        
[(0 : 1 : 0)]

Which means the only solution is the point at infinity of the projective plane, which does not concern us.
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