Can someone verify the reasoning of why these two congruences are equivalent?

Question

I've wondered why two congruences like $x\equiv81\pmod {53}$ and $x\equiv28\pmod{53}$ were equivalent.

I've come up with this proof. I am not sure if this is how you prove it though
I know that $x\equiv81\pmod{53}$ is saying that the difference between $x$ and $81$ is a multiple of 53. Representing that algebraically, we get
$\quad x - 81 = 53k$ where k is some integer
$\quad$ You can break -81 down into -53 - 28, to get
$\quad x - 53 - 28 = 53k.$
$\quad$ Adding $53$ to both sides, we get
$\quad x - 28 = 53(k+1)$
$\quad$ Making a substitution $n = k + 1$, we finally get the expression
$\quad x - 28 = 53(n)$ where n is some integer
$\quad$ Which is by definition $x\equiv28\pmod{53}$

Is this how you guys would prove this or the general way to prove this? Can you guys verify this "proof"?

Answer 1

Because $81$ and $28$ are congruent $\bmod 53$. one of the very first things that should be proved when learning modular arithmetic is that congruence is an equivalence relation. So $x\equiv 28 \iff x\equiv 81 \bmod 53$

Answer 2

Your argument is fine, but a bit longer than necessary.

Note that, in general $d\mid n \iff d\mid (n-d)$.

So $53\mid (x-28) \iff 53\mid ((x-28)-53)=x-81$.