I was reading a solution to a problem of IMO 1982.
Here is the problem
Prove that if n is a positive integer such that the equation $x^ 3 − 3xy^2 + y^ 3 = n$ has an integer solution (x, y) then it has at least three such solutions.
In the solution of Titu andreescu,He used the fact that ,If $(x,y)$ is a solution to the equation then \begin{align*}x^3 -3xy^2 +2y^3 &=(y-x)^3 -3x^2y+2x^3\\&=(y-x)^3-3(y-x)x^2+(-x)^3\ \end{align*} $\Rightarrow(y-x,-x)$is also a solution to the equation.\begin{align*}T\begin{pmatrix}x\\y\end{pmatrix} \end{align*}\begin{align*}&=\begin{pmatrix} -1 & 1\\-1 & 0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\\&=\begin{pmatrix}-x+y\\y\end{pmatrix} \end{align*} Also,\begin{align*} T^2 &= \begin{pmatrix} -1 & 1\\ -1 & 0 \end{pmatrix} \begin{pmatrix} -1 & 1\\ -1 & 0 \end{pmatrix} \\&= \begin{pmatrix} 0 & -1\\ 1 &-1 \end{pmatrix} \end{align*} under the linear transformations,$T, T^2$ on\begin{pmatrix} x\\y\end{pmatrix}, the solution of the equation $(x,y)$ proves to be valid. but how? I don't know much about linear algebra but can anyone explain to me that how these solutions are kept valid by linear transformation?
Moreover, is there any normal algebraic version of this solution?