Non-negative integer solutions to the equation $x^3-x^2+x=3y^3$

Question

Find all the non-negative integer solutions to the equation : $$x^3-x^2+x=3y^3$$

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Some thoughts.

It is easy to see that, $x=y=0$ is a trivial solution.

I realized that,

$$x(x^2-x+1)=3y^3$$

and I am not sure how can this help, however I know that

$$x\left(\frac {x^3+1}{x+1}\right)=3y^3$$ $$x(x^3+1)=3y^3(x+1)$$

I don't see anything useful here.

Suppose $x\mid 3$ or $x=3k$, I see that

$$27k^3-9k^2+3k=3y^3$$ $$9k^3-3k^2+k=y^3$$

or

$$k(9k^2-3k+1)=y^3$$

Maybe $k=q_1^3$ and $9k^2-3k+1=q_2^3$ can help here, However, I have no good reason to follow this assumption.

Is a more satisfying answer possible involving higher mathematics?

Answer 1

Assume that $(x,y)$ is a rational solution of the given equation $x^3 -x^2 +x=3y^3$. The case $x=0$ corresponds to the solution $(0,0)$. Else, let us assume $x\ne 0$. Then: $$ \left(\frac yx\right)^3-\frac 14 = \frac{4y^3-x^3}{4x^3}= \frac{\frac 43(x^3-x^2+x)-x^3}{4x^3}= \frac{\frac 13(x^3-4x^2+4x)}{4x^3}= \frac 1{12}\left(\frac{x-2}x\right)^2 \ . $$ To get an elliptic curve we multiply with $12^3$ and set $\displaystyle X=12\cdot \frac yx$, and $\displaystyle Y=12\cdot \frac {x-2}x$.

Then we obtain, using $\frac 14\cdot 12^3=432$: $$ \bbox[yellow]{\qquad Y^2 = X^3-432\ .\qquad} $$ This elliptic curve has rank zero, and the only rational points (excluding its point at infinity) are $(12,\pm 36)$:

sage: E = EllipticCurve(QQ, (0, -432))
sage: E.rank()
0
sage: E.torsion_order()
3
sage: E.torsion_points()
[(0 : 1 : 0), (12 : -36 : 1), (12 : 36 : 1)]

The possible $Y$-values are thus $\pm 36$. From $\pm36=Y=12\frac{x-2}x$, we get $\pm 3x=x-2$, so $x$ is either $-1$, or $x$ is $\frac12$. We obtain the rational solutions: $$(-1, -1)\ ,\qquad\left(\frac 12,\frac 12\right)\ .$$ Which are not solution under the wanted condition. (The first solution has negative components, the second one has non-trivial denominators.)

Answer 2

You can observe that :

$$ \begin{align}&x^3-x^2+x=3y^3\\ \iff &243(x^3-x^2+x)=(9y)^3\end{align} $$

Then let $\thinspace (a,b,c,d)\in\mathbb Z^4$, we assume that the polynomial $243(x^3-x^2+x)$ is equivalent to :

$$ \begin{align}(ax+b)^3+(cx+d)^3=(a^3+c^3)x^3+(3a^2b+3c^2d)x^2+(3ab^2+3cd^2)x+(b^3+d^3)\end{align} $$

This implies that :

$$ \begin{align} &\begin{cases}d=-b\\ a^3+c^3=243\\ b(a^2-c^2)=-81\\ b^2(a+c)=81\end{cases} \\\\ \implies &\begin{cases}\frac{b^2(a^2-c^2)^2}{b^2(a+c)}=(a-c)^2(a+c)\\ (a+c)(a^2-ac+c^2)=243\\ (a-c)^2(a+c)=81\end{cases}\\\\ \implies &\begin{cases}\frac {a^2-ac+c^2}{(a-c)^2}=\frac {a^2-ac+c^2}{a^2-2ac+c^2}=3\thinspace .\end{cases}\end{align} $$

Substituting $\thinspace u=\frac ac$, then we have :

$$ \begin{align}&\frac {u^2-u+1}{u^2-2u+1}=3\\ \implies &u\in\left\{2,\frac 12\right\}\\ \implies &a=2c \thinspace\thinspace\thinspace \text{or}\thinspace\thinspace\thinspace c=2a\thinspace .\end{align} $$

Wlog $\thinspace c=2a$. Then :

$$ \begin{align}&a^3+(2a)^3=243\\ \implies &\begin{cases}a=3,\thinspace c=6\\ b=\frac {-81}{3^2-6^2}=3\\ d=-b=-3\end{cases}\end{align} $$

Thus, we obtain the following identity :

$$ \begin{align}243(x^3-x^2+x)&=(3x+3)^3+(6x-3)^3 \end{align} $$

which means :

$$(3x+3)^3+(6x-3)^3=(9y)^3$$

Finally by Fermat's last theorem, we find the pair of $(x,y)=(0,0)$ is the only possible nonnegative integer solution .