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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.

Michael Rozenberg
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Manan
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    This doesn’t quite show why they can’t be $6n+2$ or $6n+4$, which are $2$ and $1$ modulo $3$, respectively. Fortunately, you can easily show they cannot be the case (why?). – paulinho Aug 07 '20 at 05:10
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    @paulinho As suggested by Michael in his answer below, even parity would imply divisibility by 2. – Manan Aug 07 '20 at 05:11

2 Answers2

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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.

Michael Rozenberg
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Regarding your proof, you need to consider the case where $n'$ is odd: when $n' = 2k+1$ for a positive integer $k$. When $n'$ is odd, what happens to the parity of $p$ in both cases?

Toby Mak
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  • Noted, thanks! I should have avoided an arbitrary substitution like that. – Manan Aug 07 '20 at 05:10
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    Splitting the problem into cases is an effective strategy, so if you justify your reasoning this way, your proof is fine. However, in hindsight, Michael Rozenberg's proof is much shorter. – Toby Mak Aug 07 '20 at 05:12