Transivity proof : $y^2$ divides by $2x$ Given a relation $R=\{(x,y) : y^2 \hbox{ divides } 2x\}$
For $x,y$ natural numbers.
How to go about proving transitivity of the set? 
I have tried many ways.
Or maybe I am mistaken and this is not true?
 A: Since the question isn't very clear (see @vadim123's comment), I'll just do both situations:
Case 1: $R = \{(x,y) \mid \exists k \in \mathbb{N}: k \cdot y^2 = 2x\}$. 
Suppose $(x,y) \in R$ and $(y,z) \in R$. Then there exists some $k \in \mathbb{N}$ such that $ky^2 = 2x$ and there exists some $l \in \mathbb{N}$ such that $lz^2 = 2y$.
Take $n = \frac{kly}{2}$. Since $ky^2 = 2x$, at least one of $k$ and $y$ has a factor $2$, hence $n \in \mathbb{N}$. Now, $nz^2 = \frac{ky}{2}(lz^2) = \frac{ky}{2} \cdot (2y) = ky^2 = 2x$. Hence $(x,z) \in R$.
Case 2: $R = \{(x,y) \mid \exists k \in \mathbb{N}: k \cdot 2x = y^2\}$. 
Note that $(8,4) \in R$ since $1 \cdot (2\cdot 8) = 16 = 4^2$; and $(32,8) \in R$ since $1 \cdot (2 \cdot 32) = 64 = 8^2$. Transitivity would require that $(32,4) \in R$: there must exist some $k \in \mathbb{N}$ such that $k \cdot (2 \cdot 32) = 64k = 16 = 4^2$. Clearly this is impossible, so $R$ is not transitive in this case.
A: The key idea is that $\ x\sim y\,\Rightarrow\, y\mid x,\,$ proved below, writing $R$ infix as $\,\sim$
Note $\ x\sim y\,\Rightarrow\,2x = ny^2\,\Rightarrow \left\{ \begin{align}&\color{#0a0}{2\mid n}\,\Rightarrow\, x = \color{#0a0}{n/2}\,y^2\\ {\rm or}\ \  &\color{#c0f}{2\mid y}\,\Rightarrow\, x = ny\,\color{#c0f}{y/2}\end{align}\right\}\, $  so $\ \color{#c00}{y\mid x}\,$ 
Thus $\ y\sim z\,\Rightarrow\,z^2\mid 2\color{#c00}y\mid 2\color{#c00}x,\ $ so  $\ x\sim z.\ $  Thus $\,\sim\,$ is transitive.
