Solve $y^3=x^{3}+8x^{2}-6x+8$ for positive integers $x, y $

Question

How we can solve this equation : $$y^3=x^{3}+8x^{2}-6x+8$$ for positive integers $x, y $? I tried to factories the $L. H. S $ but I couldn't complete the solution. How I can solve it? Thanks.

Answer 1

Since we have$$\begin{align}&(x+2)^3\lt x^3+8x^2-6x+8\lt (x+3)^3\\&\iff x^2-9x\gt 0\quad\text{and}\quad x^2+33x+19\gt 0\\&\iff x\gt 9\end{align}$$ we know that if $x\gt 9$, then there is no such positive integer $y$. So, we have $x\le 9$.

Now, all we need is to check if each of $x=1,2,\cdots,9$ is sufficient.

Thus, we know that $(x,y)=(9,11)$ is the only solution.

Answer 2

It's clear by inspection that $x$ and $y$ must have the same parity, so let's write $y=x+2k$. The equation reduces to a quadratic in $x$:

$$(4-3k)x^2-3(1+2k^2)x+4(1-k^3)=0$$

The discriminant of this quadratic equation is

$$\begin{align} \Delta(k)&=9(1+2k^2)^2-16(4-3k)(1-k^3)\\ &=-(12k^4-64k^3-36k^2-48k+55) \end{align}$$

For the quadratic to have any real (much less integer) roots, $\Delta(k)$ must be non-negative. This clearly restricts $k$ to a finite interval, which turns out to be $1\le k\le5$. (It's easy to see $\Delta(k)\lt0$ if $k\le0$. The verification that $\Delta(k)\lt0$ if $k\ge6$ is little messy, so I'm omitting it. However, you can, if you like, skip from here to the "added later" section below.)

For the quadratic to have any integers roots, $\Delta(k)$ must be a square, so at this point it's easiest to just calculate:

$$\begin{align} \Delta(1)&=81\\ \Delta(2)&=505\\ \Delta(3)&=1169\\ \Delta(4)&=1737\\ \Delta(5)&=1585\\ \end{align}$$

The only square here is when $k=1$, so we see the quadratic

$$x^2-9x=0$$

which has roots $x=0$ and $x=9$. Thus the only solution in positive integers is $(x,y)=(9,11)$, and the only other solution in integers is $(x,y)=(0,2)$.

Remark: This solution and mathlove's are quite different, but they both lead to computing a small number of cases (nine in mathlove's, five here).

Added later: I got to wondering if there's some easy way to avoid doing a case-by-case computational check. It turns out there is, at least if you restrict to $x\gt0$ (which is specified by the OP).

Starting from the observation that $k$ must be positive in order for $\Delta(k)$ to be non-negative, let's go back to the quadratic in $x$ and rewrite it as a cubic in $k$:

$$4k^3+6xk^2+3x^2k-(4x^2-3x+4)=0$$

Thinking of the left hand side as a function $f(k)$ (holding $x$ constant), note that

$$f'(k)=12k^2+12xk+3x^2=3(2k+x)^2\ge0$$

Consequently, if $k\ge2$,

$$\begin{align} 4k^3+6xk^2+3x^2k-(4x^2-3x+4)&\ge32+24x+6x^2-(4x^2-3x+4)\\ &=2x^2+27x+28\\ &\gt0\qquad\text{if }x\gt0 \end{align}$$

This leaves $k=1$ as the only possible (integer) value that allows for a positive integer value of $x$, which turns out to be $x=9$ (hence $y=11$).