Why be a square when you can be a cube?

Number Theory Level 5

Solve the following equation over integral values of $y,x$ :

$y^3=x^3+8x^2-6x+8$

Details and assumptions:

Please input your answer as the sum of all possible values for $(x,y)$ .As an explicit example,if the ordered pair $(1,2)$ is an answer and is the only one,input $3$ but if $(2,3)$ is also an answer, input $8$
$x,y \in \mathbb {Z}^+$

The answer is 20.

4 solutions

Mathh Mathh
May 15, 2015

I'll solve it over integers instead. Solving quadratic inequalities gives:

$(x+2)^3<x^3+8x^2-6x+8$ for all $x$ except when $x\in[0,9]$ .

$x^3+8x^2-6x+8<(x+3)^3$ for all $x$ except when $x\in[-32,-1]$ .

For those exceptions when $x\in [-32,9]$ (otherwise $x^3+8x^2-6x+8$ is between two consecutive cubes and can't be a cube) I wrote a C++ program, but they can be checked manually. All solutions over integers are $(x,y)=(0,2),(9,11)$ . Of course, over $\mathbb Z^+$ checking only $[1,9]$ without a program is quick.

#include <iostream>
#include <cmath>

using namespace std;

int main() {
int x;
for (x=-32;x<=9;x++)
 if (cbrt(x*x*x+8*x*x-6*x+8)==floor(cbrt(x*x*x+8*x*x-6*x+8)))
  cout << "(x,y)=(" << x << "," << cbrt(x*x*x+8*x*x-6*x+8) << ").\n";
return 0; }

Although I don't quite understand what you wrote but I always love solutions with some CS used in them.Great job!

Arian Tashakkor - 6 years, 1 month ago

My code uses this nice property I thought of: $a$ is an integer iff $a=\lfloor a\rfloor$ . When checking if $k$ is a cube, it is significantly more efficient to check if $\sqrt[3]{k}=\lfloor \sqrt[3]{k}\rfloor$ than e.g. to go through all $i$ with $i\in [1,\sqrt[3]{k}]$ and checking if $i^3=k$ .

mathh mathh - 6 years, 1 month ago

I think you got me wrong xD.What I meant was that I don't make head or tails of coding in any language of any kind.But I did get what you meant in your last comment tho.We used the same approach for writing algorithms (yes I'm a new learner :D) when we wanted to see if an integer $n$ is divisible by another integer $m$ by checking that $\frac{n}{m}=\lfloor \frac{n}{m} \rfloor$

Arian Tashakkor - 6 years, 1 month ago

Des O Carroll
May 15, 2015

Let y=x+a and so x^3 +3ax^2+3a^2x+a^3=x^3 +8x^2-6x=8

giving (3a-8)x^2+(3a^2+6)x +(a^3-8)=0

If a=1 no integral solutions

If a=2 we get x=0(which we discount) or x=9 and y=11

Now if the equation px^2 +qx +r=0 must have at least one positive solution then pr<0

and if a>2 by inspection we see that (3a-8)(a^3-8)>0

therefore only solution is x=9 y=11

Niranjan Khanderia
May 18, 2015

I had written a solution. It has vanished!!!!I am checking. Well I write it again. $y^3=x^3+8x^2-6x+8,~~\\\implies~y^3-x^3=8x^2-6x+8=2*(4x^2-3x+4)..(1)\\But~y^3-x^3 =(y-x)(y^2+x^2+xy),..(2)~and~x,y \in \mathbb {Z}^+.\\\therefore~(y-x)=2 ~is~a~possibility.~~ Trying~this~\\from~(1)~and~(2),~~4x^2-3x+4=(x+2)^2+x^2+x*(x+2)\\Solving~x^2=9x. ~~~x\neq 0,\\ \therefore x=9~and ~y=11,~satisfies ~the ~equation. ~answer=9+11=~~\color{#D61F06}{20}\\$ $\color{#EC7300}{OR::-~~~y~ must~ be~=x+integer~a.\\ (x+a)^3=x^3+3*x*a*(x+a)+a^3, ~\therefore~a^3=8, \therefore~ try~a=2\\\therefore~6x^2+12x=8x^2-6x. ~x\neq0~\therefore~2x=18,~x=9....and ~y=11..}\\~~\\$ $\color{#3D99F6}{\text{For those using calculator,(like me) after 0 STR x:1+x STR x: use }\\ ( x^3+8x^2-6x+8 )^{\frac 1 3 }~\text{ STOP when the return is an integer. } }$

Aakash Khandelwal
May 15, 2015

y^3=(x+2)^3+2x(x-9)

To be a perfect cube x(x-9)=0

i.e. x=0 or x=9 but since x>0 hence x=9 only.

therefore required only ordered pair=(11,9)

answer=20

Moderator note:

Your reasoning is flawed. If we rearrange your first equation, we would get $y^3 - (x+2)^3 = 2x(x- 9 )$ , which means it's a difference of two perfect cubes. It does not necessarily mean that this difference of perfect cubes must be 0.