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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.


marked as duplicate by Joel Reyes Noche, Misha Lavrov, Robert Soupe, Namaste, steven gregory Apr 5 '18 at 19:22

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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.

  • $\begingroup$ Nice Solution. But I have a question. How to invent $$\begin{align}&(x+2)^3\lt x^3+8x^2-6x+8\lt (x+3)^3\\&\iff x^2-9x\gt 0\quad\text{and}\quad x^2+33x+19\gt 0\\&\iff x\gt 9\end{align}$$ in exam setup. Is it a result of experience gained from massive number of problem solving or you have used some other insight to arrive it which in final solution you have omitted? In short can you please elaborate your thought process. Thanks in advance. $\endgroup$ – rugi Mar 23 '16 at 6:31
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    $\begingroup$ @rugi: I just wanted to get integers $a,b$ such that $(x+a)^3\lt x^3+8x^2-6x+8\lt (x+a+1)^3$ is true for all $x\gt b$. Then, noting that the coefficient of $x^2$ is $8$ in the middle polynomial, we can see that $a$ has to be $2$, and so on. I hope this helps. $\endgroup$ – mathlove Mar 23 '16 at 6:39
  • $\begingroup$ Thanks. It is now clearer. But why to think of restricting between $(a)$ and $(a+1)$. I mean to say why not $(a)$ and $(a+k)$ where $k$ is any integer? I hope I am not asking something really silly. $\endgroup$ – rugi Mar 23 '16 at 6:43
  • $\begingroup$ @rugi: Because $(x+a)^3$ and $(x+a+1)^3$ are consecutive cubes, which is the key of this answer. As we can have $(x+2)^3\lt x^3+8x^2-6x+8\lt (x+3)^3$ for all $x\gt 9$ where $(x+2)^3$ and $(x+3)^3$ are consecutive cubes, if $x\gt 9$, then $x^3+8x^2-6x+8$ cannot be a cube, which means that there is no integer $y$ such that $y^3=x^3+8x^2-6x+8$ as written in the answer. $\endgroup$ – mathlove Mar 23 '16 at 6:47

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 $x$:


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


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 $k$:


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


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 $y=11$).

  • $\begingroup$ Nice solution sir. but I think we will refuse t$(0, 2) $ because we want to solve the equation for $\large{positive}$ integers $x, y $... or I missed something? $\endgroup$ – user354387 Sep 15 '15 at 15:34
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    $\begingroup$ @user256952, no, you didn't miss anything. It was really just a remark, in case anyone wondered if the equation had any non-positive integer solutions. (For some problems, restricting to positive integers is crucial, in terms of finding a simple solution, while for other problems it isn't; this problem is of the latter variety.) $\endgroup$ – Barry Cipra Sep 15 '15 at 16:55
  • $\begingroup$ I love this solution because it is more illuminating than the previous answer. $\endgroup$ – rugi Mar 23 '16 at 6:34

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