Prove that $x\mathrel Ry\iff4\mid(3x+y)$ is Transitive

Question

Let $x,y,z \in\mathbb Z$ and suppose that $x\mathrel Ry$ and $y\mathrel Rz$. Therefore $4\mid(3x+y)$ and $4\mid(3y+z)$.

So there exists $k, l \in\mathbb Z$ such that $4k=3x+y$ and $4l=3y+z$. Add these equations together to get $4k+4l=3x+4y+z$. Subtract $4y$ from both sides to get $4(k+l-y)=3x+z$

Note that $k+l-y \in\mathbb Z$, because $k,l,y \in\mathbb Z$ and $\mathbb Z$ is closed under addition and subtraction.

So $4\mid(3x+z)$.

So $x\mathrel Rz$.

Answer 1

Observe that your relation can be defined easily and equivalently by

$$xRy \;\;\iff \;\; 4|(y-x)$$

So,

$$xRy \text{ and } yRz \;\implies $$

$$4|(y-x) \text{ and }\; 4|(z-y)\, \implies$$

$$4|\Bigl((y-x)+(z-y)\Bigr)\; \implies$$

$$4|(z-x)\;\; \implies xRz.$$