Proof-verification: Any prime $p>3$ can be expressed in the form $(6n+1)$ or $(6n+5)$ I have already seen answers to this question, but I would like to get my own proof verified.

Proposition: Any prime $p>3$ can be expressed in the form $(6n+1)$ or $(6n+5)$.


Definition: A prime $p$ is a natural number greater than $1$ which is divisible only by $1$ and itself. Thus, no prime $p>3$ is divisible by $3$.


Proof: $$p=3n'+r\;\mathrm{where}\;0<r<3$$
$r\neq 0,3$ because stating otherwise would imply divisibility by 3. The above statement is equivalent to saying: $$(p=3n'+1)\,\mathrm{or}\,(p=3n'+2)$$ Further, $(3n'+2)\equiv5\mod3\iff(3n'+5)\equiv2\mod3$. Thus, $$p=(3n'+1)\;\mathrm{or}\;p=(3n'+5)$$Substituting $n'=2n$, we get: $$p=(6n+1)\;\mathrm{or}\;p=(6n+5)$$ Which completes the proof.

I'm sceptical of the step where I used congruence to switch $2$ and $5$. I would like to know the flaws in my proof, and measures that I can take to rectify them.

Let $n$ be a non-negative integer number.
We see that $6n$ is divisible by $6$, $6n+2=2(3n+1)$ is divisible by $2$,
$6n+3=3(2n+1)$ is divisible by $3$ and $6n+4=2(3n+2)$ is divisible by $2$.
Id est, may be $p=6n+1$ or $p=6n+5$ only.
