Find all $x,y$ such that $x^3+5y^3=(a^3+5b^3)^3$.

Question

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.

Answer 1

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.

Answer 2

As noted in the other answer, this is a question about rational points on elliptic curves:

Claim: There are no solutions to $x^{3}+y^{3} = 5$ with $x,y \in \mathbb{Q}$.

This follows from the following Proposition:

Proposition: Set $R := \mathbb{Z}[t]/(t^{2}+t+1)$. Let $A,B,C$ be elements of $R_{\mathbb{Q}}$ such that \begin{align*} A+B+C = 0 \quad\text{ and }\quad ABC = 5w^{3} \end{align*} for some $w \in R_{\mathbb{Q}}$. Then two of $A,B,C$ must be equal.

Proof of Proposition: See this post by Paul Monsky (with "2" instead of "5"); I write some more details here. It seems this argument is called "3-descent" and is usually attributed to Sylvester, Pepin, Lucas in the literature. I think we can replace "5" with any prime $p \in \mathbb{Z}$ such that $p$ remains prime in $R$.

We will use without proof that $R$ is a UFD. Let $\xi \in R$ be the class of $t$. There is a norm map \begin{align*} |\;\;| : R \to \mathbb{Z} \end{align*} sending \begin{align*} a+b\xi \mapsto \det(a\begin{bmatrix} 1 & \\ & 1 \end{bmatrix} + b\begin{bmatrix} & -1 \\ 1 & -1 \end{bmatrix}) = a^{2}-ab+b^{2} = (a+b\xi)(a+b\xi^{2}) \end{align*} for all $a,b\in \mathbb{Z}$. (We always have $a^{2}-ab+b^{2} = (a-\frac{1}{2}b)^{2}+\frac{3}{4} b^{2} \ge 0$.) The norm is multiplicative, satisfies the triangle inequality, and $a+b\xi$ is a unit of $R$ if and only if $|a+b\xi|$ is a unit of $\mathbb{Z}$ (i.e. $|a+b\xi| = 1$).

If there's a counterexample to the Proposition, there's one with $A,B,C$ in $R$ by clearing denominators (then $w \in R$ also); take such a counterexample with \begin{align*} d = \max(|A|,|B|,|C|) \end{align*} as small as possible. Here necessarily $d > 1$ (otherwise each $A,B,C$ would be either $0$ or a unit); in fact $d > 2$ since $2$ is not a norm from $R$ (the equation $(2a-b)^{2}+3b^{2} = 8$ has no integer solutions). Since $A+B+C = 0$ and $(A,B,C)$ was assumed minimal, we have that $A,B,C$ are pairwise coprime. Note that $5$ is a prime in $R$ since $R/(5)R \simeq \mathbb{F}_{5}[t]/(t^{2}+t+1)$ is a field. Note also that the only units of $R$ are $\pm \xi^{i}$ for $i \in \{0,1,2\}$. By unique factorization in $R$, there exist $r,s,t \in R$ such that \begin{align*} A = 5\xi^{i}r^{3} \quad\text{ and }\quad B = \xi^{j}s^{3} \quad\text{ and }\quad C = \xi^{k}t^{3} \end{align*} for some $i,j,k \in \{0,1,2\}$. Here $w^{3} = \xi^{i+j+k}(rst)^{3}$ so $i+j+k \equiv 0 \pmod{3}$; we may thus assume that $i=j=k=0$, so that \begin{align*} A = 5r^{3} \quad\text{ and }\quad B = s^{3} \quad\text{ and }\quad C = t^{3} \end{align*} in $R$. Consider \begin{align*} A' := s+t \quad,\quad B' := \xi^{1}s+\xi^{2}t \quad,\quad C' := \xi^{2}s+\xi^{1}t \end{align*} in $R$; then $A'+B'+C' = 0$ and $A'B'C' = s^{3}+t^{3} = -5r^{3}$; it remains to show that they are distinct. We have $\xi^{1}s+\xi^{2}t \ne \xi^{2}s+\xi^{1}t$ since $\xi^{1}-\xi^{2}$ is a nonzerodivisor in $R$ (since $t-1$ does not divide $t^{2}+t+1$) and $s,t$ are distinct (since $B,C$ are distinct). We have $s+t \ne \xi^{1}s+\xi^{2}t$ (since otherwise $(1-\xi)s = (\xi^{2}-1)t$ implies $-s = (\xi+1)t$ implies $s = \xi^{2}t$ but $s,t$ are coprime in $R$). By symmetry, we have $s+t \ne \xi^{2}s+\xi^{1}t$.

We have $|\xi^{i}s+\xi^{j}t| \le |s|+|t| = |B|^{1/3}+|C|^{1/3} \le 2d^{1/3}$, so $(A',B',C')$ is a new counterexample with $\max(|A'|,|B'|,|C'|) \le 2d^{1/3} < d$.