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Possible Duplicate:

Is that true that all the prime numbers are of the form $6m \pm 1$?

Q. Why is it that all primes greater than 3 are either 1 or -1 modulo 6?

Does it suffice to argue as follows:

- A conjecture about prime numbers
- Proving $\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1$
- Ternary Quadratic Forms
- Find $n$ such that $n/2$ is a square, $n/3$ is a cube, and $n/5$ a fifth power
- Proof strategy - Stirling numbers formula
- Why is $a$ and $b$ coprime if $a\equiv 1 \pmod{b}$?

Let $p$ be a prime. $p>3 \Rightarrow 3$ does not divide $p$. Clearly $2$ does not divide $p$ either, and so 6 does not divide p.

Now, $p$ is odd, and so $p$ is either 1,3 or 5 modulo 6. However, if $p$ were 3 (mod6), that would give us that $3$ divides $p$, which is a contradiction.

As such, we conclude that $p$ is either 1 or 5 (=-1) mod 6

- Can n! be a perfect square when n is an integer greater than 1?
- Divisibility question
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- What is complete induction, by example? $4(9^n) + 3(2^n)$ is divisible by 7 for all $n>0$
- find remainder of a number
- Does $k=9018009$ have a friend?
- Find the sum of all quadratic residues modulo $p$ where $p \equiv 1 \pmod{4}$
- When is a rational number a sum of three squares?
- Solution to exponential congruence
- How to show that $\displaystyle = \frac{abc}{(ab,bc,ca)}$ without prime factorization?

Yes. Generally if $\rm\:p\:$ is prime, then modulo $\rm\,2p,\,$ any prime $\rm\:q\ne 2,p\:$ must lie in one of the $\rm\:\phi(2p) = \phi(p) = p\!-\!1\:$ residue classes that are coprime to $2$ and $\rm p,$ i.e. all odd residue classes excluding $\rm p,\:$ viz. $\rm\:1,3,5,\ldots,\hat p,\ldots,2p\!-\!1.\:$ Indeed, integers in other classes are divisible by $2$ or $\rm p\:$ hence, if prime, must be $2$ or $\rm p,\:$ resp. More succinctly, exploiting negation reflection symmetry: $\rm\: q\equiv \pm\{1,3,5,\cdots,p\!-\!2\}\ \ (mod\ 2\:\!p),\ $ e.g. $\rm\,\ q\equiv \pm 1\ \ (mod\ 6),\:$ $\,\rm q\equiv \pm\{1,3\}\ \ (mod\ 10),\:$ $\,\rm q\equiv\pm \{1,3,5\}\ \ (mod\ 14),\:$ etc.

Generally, if $\rm\,q\,$ is *any integer* coprime to $\rm\:m\:$ then its remainder mod $\rm\:m\:$ lies in one of the $\rm\:\phi(m)\:$ residue classes coprime to $\rm\:m,\:$ where $\phi$ is the Euler totient function.

We have $6| 6n$, $2| 6n+ 2$, $3|6n+3$, and $2|6n + 4$ for all integers $n$. That leaves the $6n+1$s and $6n+5$s as possible primes, save for 3 and 2.

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