Thread:

1. Every prime > 3 is congruent to \pm 1 mod 6

Every prime > 3 is congruent to $\pm 1 \ \mbox{mod 6}$

Induction maybe?

$5\equiv\pm 1\ \mbox{(mod 6)}$ This isn't true though

2. any number can be

0,1,2,3,4,5 mod 6

because the number is prime - 0,2,3,4 can be ruled out

3. $5\equiv -1\ \mbox{(mod 6)},$ since $5-(-1)=6\times 1$.

You could try arguing along these lines: all primes greater than 3 are odd. Therefore, they must be congruent to either -1, 1, or 3 mod 6. (5 and -1 are equivalent). However, any number greater than 3 that is congruent to 3 mod 6 would be divisible by 3, and hence not prime. Therefore, all primes greater than 3 are congruent to 1 or -1 mod 6.

[EDIT]: This is essentially the same as what aman_cc said.

4. Originally Posted by dwsmith
Every prime > 3 is congruent to $\pm 1 \ \mbox{mod 6}$

Induction maybe?

$5\equiv\pm 1\ \mbox{(mod 6)}$ This isn't true though
You can generalize and consider integers $a$ that are relatively prime to $6$ and the result will follow, in particular, to any prime $p>3$.

Let $a=6q+r$, with $0\leq r<6$. Then $gcd(a,6)=gcd(6,r)$, so we need only to look at values $r$ such that $gcd(6,r)=1$. These are $r=1,5$.

Now, $a\equiv 1(mod\ 6)$,or $a\equiv 5\equiv -1(mod\ 6)$.

5. Keep it simple ... I assume that $p > 3$ in the whole post by the way ... since the fact $3 | 3$ does not contradict the primality of $p$ ... just adjust what is below as you wish ...

Let $p \equiv x \pmod{n} \Rightarrow p = kn + x$, $k \in \mathbb{Z}$, $x \in \mathbb{Z}/n\mathbb{Z}$
Then it follows that $\gcd{(n, x)} | p$
But if $p$ is prime we must have $\gcd{(n, x)} = 1$ (otherwise there is a contradiction with respect to the definition of a prime number)

Applying this to the current problem with $n = 6$, the only $x$ that are coprime to $6$ (in the congruence ring $\mathbb{Z}/6\mathbb{Z}$ of course) are, surprisingly enough, $1$ and $5$ (which turns out to be congruent to $-1 \pmod{6}$).

It follows that if $p$ is indeed prime then it must be congruent to $\pm 1 \pmod{6}$

...

Interestingly, it is also possible to prove that this implies that $\gcd{(k, n)} = 1$ for any prime $p$ ... This gives me an idea, I'll be right back !

6. To Bacterius

gcd(n,x) = 1 or p

i.e. p = p mod p^2, so gcd(p^2,p) = p, and p|p

But since n is 6, then we can say that gcd(n,x) should be 1

7. I don't understand what you mean, can you develop ?

8. I was just pointing out that gcd(n,x) isn't necessarily 1, it could be p also, at least in general ... for this particular case, since n is 6, then gcd(n,x) would be 1 ...
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