Can someone write a solution?

Factorize LHS as Difference of Squares, to obtain the following expression,

$8(x^2-2x+2)(1-x)=y^3 \;$ $\quad$

As it can be seen, for $x,y \in \mathbb{W}$ , $\quad$

$x$ can only equal $1$ so that the LHS stays non-negative ; and consequently $y=0$ . $\quad$

$\therefore (1,0)$ is the only solution.

Firts we factorize $(x-2)^2 - x^4$

$(x-2)^2 - x^4$

= $(x-2)^2 + x^2)((x-2)^2 - x^2)$

= $(x-2+x)(x-2-x)((x-2)^2 + x^2)$

= $(2x-2)(-2)(2x^2-2x+4)$

= $(-8)(x-1)(x^2-2x+2)$

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

Since -8 is already a cube then that means that $(x-1)((x-1)^2 +1)$ must also be a cube hence

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

since $(x-1)$ and $((x-1)^2 +1)$ $\in \mathbb{Z}$ We can split this into 4 cases:

Case 1

$x-1=k$

$(x-1)^2+1=k^2$

Case 2

$x-1=k^2$

$(x-1)^2+1=k$

Case 3

$x-1=1$

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

Case 4

$x-1=k^3$

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

Solving the equations we find that the only possible solution is $x=1$ and $k=0$ :

$(-8(k^3)=y^3$

$(0)=y^3$

$y=0$

So the only pair that satisfies this equation is $(0;1)$

$(x-2)^4-x^4=y^3$ We can factor this to give $(-2)^3(x-1)((x-1)^2+1)=y^3$

Since we have $(-2)^3$ on the left, the other factors must also give a cube, so $(x-1)((x-1)^2+1)=z^3$

But $x-1$ and $(x-1)^2+1$ are clearly coprime, and the only way for coprime values to multiply to give a cube is if they are both cubes themselves. This tells us that

$x-1=n^3$

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

Substituting $x-1=n^3$ into the second equation gives

$n^6+1=m^3$

And so

$1=m^3-n^6=(m-n^2)(m^2+mn^2+n^4)$

Since the two factors on the right are both integers, they must be $\pm1$ .

However $m>0$ by our definition of $m$ , which tells us the rightmost factor is positive and so both factors equal $1$ , but since $m>0$ this can only only happen when $m=1, n=0$ . Therefore the other solution is

$x-1=0\implies x=1$ .

(x-2)^4 - x^4 = y^3 Or, y^3 = -2 (2x-2)(2x^2 - 4x +4) =-8 (x-1)(x^2 - 2x + 2) =(-2)^3×[(x-1)(x^2 - 2x + 2)] Therefore, for x and y to be whole number (x-1)(x^2 - 2x + 2) has to be perfect cube (x-1)(x^2 - 2x + 2)=(x-1)^3 + (x-1) Therefore (x-1)(x^2 - 2x + 2) to be whole cube x-1 has to be equal to zero as (x-1)^3 is a perfect cube. Therefore x-1=0 x=1 This the only solution for which (x,y) are whole numbers. Therefore there is only 1 solution. (x,y) = (1,0)