Integer solutions to $2x - 3y = z$ and $2y - 3z = x$

Question

Are there infinitely many integer solutions ($x, y, z$) to the pair of equations $2x-3y=z$ and $2y-3z=x$?

For instance, $(1, 1, 1)$, $(11, 7, 1)$, and $(-11, -7, -1)$ are solutions. Any help on how to find solutions in general? Thanks!

Answer 1

Hint $ $ By Cramer's rule, since the determinant $= 1, $ we have

$$\begin{align} &\begin{bmatrix} 1 & -2 \\\\ 2 & -3 \end{bmatrix} \begin{bmatrix} x \\\\ y \end{bmatrix} = \begin{bmatrix} -3z \\\\ z\end{bmatrix}\\ \\ \iff\ \begin{bmatrix} x \\\\ y\end{bmatrix} =\, &\begin{bmatrix} -3 & 2 \\\\ -2 & 1 \end{bmatrix} \begin{bmatrix} -3z \\\\ z \end{bmatrix} = \begin{bmatrix} 11z \\\\ 7z\end{bmatrix} \end{align}$$

Answer 2

Note that $(1,1,1)$ does not appear to be a valid solution.

$\begin{cases} 2x-3y=z\\ 2y-3z=x \end{cases}\iff \begin{cases} 2(2y-3z)-3y=z\\ x=2y-3z \end{cases}\iff \begin{cases} y=7z\\ x=14z-3z=11z \end{cases}$

So the solutions are $(11,7,1)\times k$ with $k\in\mathbb Z$.

Answer 3

$2x - 3y = z$ and $2y - 3z = x$ means

$2y - 3(2x-3y)= x$

$11y - 7x = 0$

So $y = \frac {7x}{11}$. So if $x$ is an integer it must be a multiple of $11$. And if $x = 11a$ then $y = 7a$.

$z = 2x - 3(\frac {7x}{11}) = \frac {x}{11}$.

So any $(11a, 7a, a)$ for $a \in \mathbb Z$ will be a solution.

$(1,1,1)$ is not a solution as $2(1) - 3(1) \ne 1$.