Proving this relation is an equivalence

Question

For $\sim$ defined on $\mathbb{Z}$ by
$$x\sim y \iff m^2 x = n^2 y$$ for some positive integers $m,n$.

I'm not sure if the transitive property is properly proved here..

Suppose $x,y,z \in \mathbb{Z}$ with $x \sim y$ and $y \sim z$.

Then, for some positive integers $m,n,k,l$,
$$m^2 x = n^2 y \qquad \text{and} \qquad k^2 y = l^2 z$$

$$y = \frac{l^2}{k^2}z$$
so $$m^2 x = \frac{n^2 l^2}{k^2}z$$
but the coefficient of $z$ is not necessarily an integer....
Am I supposed to set $k:=1$ or something?

Answer 1

$$m^2 x = n^2 y \qquad \text{and} \qquad k^2 y = l^2 z$$

$$k^2(m^2x) = k^2(n^2y)=n^2(k^2y)=n^2(l^2z)$$

That is $$(km)^2x=(nl)^2z$$

Note that $km$ and $nl$ are both positive integer.

Hence $x \sim z$.

Answer 2

Generally let $\ x\sim y \iff ax = by\,$ for some $\,a,b\in M,\,$
where $M\subset \Bbb C$ is closed under $\rm\color{#0a0}{multiplication}$. Then

$\qquad\qquad\begin{align} &\ \ \overbrace{ax =\ \ by}^{\Large x\ \sim\ y}\\ \Rightarrow\ &cax = \color{#c00}{cb}y\\ \phantom{1} \end{align}\ $ $\begin{align} &\overbrace{\ \,cy =\ dz}^{\Large y\ \sim\ z}\\ &\color{#c00}{bc}y = bdz\,\ \Rightarrow\ \underbrace{\color{#0a0}{ca}x = \color{#0a0}{bd} z}_{\Large x\ \sim\ z}\end{align}$

where $\,\color{#0a0}{ca,bd} \in M$ since $M$ is closed under $\rm\color{#0a0}{multiplication}$. Thus $\,\sim\,$ is transitive.

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Remark $ $ We multiplied the 1st equation by $\,\color{#c00}c\,$ and the 2nd by $\,\color{#c00}b\,$ in order to unify (or overlap) the coefficients of $\,y,\,$ so that we could deduce transitivity of $\,\sim\,$ via transitivity of $\,=\,$.

Such unification or overlapping is ubiquitous method of deriving consequences of equations and axioms - often employed in term-rewriting and equational / axiomatic inference systems. See this answer for further discussion.