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Prove that every prime number greater than 3 is either one more or one less than a multiple of $6$. (Hint: Consider the contrapositive by cases.)

I tried this problem using contrapositive but It is not sure, As I'm new to this subject!!

Here what I tried,

Firstly I divide this as $P\rightarrow Q$ here,

if $\forall p$ such that $p>3$ is prime then $p+1$ or $p-1$ multiple of 6

contrapositive of this statement is $\lnot Q\rightarrow \lnot P$,

if $p+1$ is not multiple of 6 and $p-1$ is not multiple of 6 then there is no prime number $p>3$.

So I assume $p+1$ is not multiple of 6 and $p-1$ is not multiple of 6 it implies

$ p+1=3(mod6)$ and $ p-1=1(mod6)$ $\rightarrow p=2(mod6)$

$ p+1=4(mod6)$ and $ p-1=2(mod6)$ $\rightarrow p=3(mod6)$

$ p+1=5(mod6)$ and $ p-1=3(mod6)$ $\rightarrow p=4(mod6)$

from these three I said there is no prime number $p>3$

Can anyone please tell me is this correct If wrong give me hint to prove!!

Théophile
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ALMEra
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    This is a very common question. The solution can be expressed much more succinctly; try to simplify your reasoning. For example: Is that true that all the prime numbers are of the form $6m \pm 1$? – Théophile Oct 26 '20 at 22:55
  • @Théophile I don't think these are duplicates. There should be an additional proof verification tag on this post. – Shubham Johri Oct 26 '20 at 22:56
  • Generally, please search first before posing questions. Then after reading prior proofs, if you still have doubts then please ask a question about the specific part of the proof you have a doubt about. If your proof is the same as a prior proof then first ask there (to help improve that answer). There are many tens (if not hundreds) of prior answers on this which are easy to locate by search. – Bill Dubuque Oct 26 '20 at 23:50

2 Answers2

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Doing a proof by contrapositive isn't the same as doing a proof by contradiction. Here you're supposed to assume that a given integer $x$ is not $\equiv 1\pmod 6$ and not $\equiv 5 \pmod 6$ and then conclude that $x$ is not a prime number. There are four cases to consider, and you have already spelled out the logic to take it from here.

In essence, because you're trying to prove the statement "if $x$ is prime then $x\equiv 1\pmod 6$ or $x\equiv 5 \pmod 6$," you can do so by proving that "if $x\not\equiv 1\pmod 6$ and $x\not\equiv 5\pmod 6$, then $x$ is not prime." This should not lead to any contradictions, e.g., "there are no primes larger than 3."

marcelgoh
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The number should be in either of the forms $4k+1$ or $4k+3$. If $k$ is divisible by $3$ then it must be the former which has the form $4(3m)+1=6(2m)+1$. If not it is the latter and it is either $3m+1$ or $3m+2$. For the first case we have: $4(3m+1)+1=12m+5=6(2m+1)-1$ and in the second case we have: $4(3m+2)+3=12m+11=6(2m+2)-1$.

am301
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