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

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

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

    Induction maybe?

    $\displaystyle 5\equiv\pm 1\ \mbox{(mod 6)}$ This isn't true though
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  2. #2
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    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
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    $\displaystyle 5\equiv -1\ \mbox{(mod 6)},$ since $\displaystyle 5-(-1)=6\times 1$.

    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.
    Last edited by Ackbeet; Jul 8th 2010 at 04:43 AM. Reason: Duplication.
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    Quote Originally Posted by dwsmith View Post
    Every prime > 3 is congruent to $\displaystyle \pm 1 \ \mbox{mod 6}$

    Induction maybe?

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

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

    Now, $\displaystyle a\equiv 1(mod\ 6)$,or $\displaystyle a\equiv 5\equiv -1(mod\ 6)$.
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    Keep it simple ... I assume that $\displaystyle p > 3$ in the whole post by the way ... since the fact $\displaystyle 3 | 3$ does not contradict the primality of $\displaystyle p$ ... just adjust what is below as you wish ...

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

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

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

    ...

    Interestingly, it is also possible to prove that this implies that $\displaystyle \gcd{(k, n)} = 1$ for any prime $\displaystyle p$ ... This gives me an idea, I'll be right back !
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    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
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  7. #7
    Super Member Bacterius's Avatar
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    I don't understand what you mean, can you develop ?
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  8. #8
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    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 ...
    Last edited by Bingk; Jul 16th 2010 at 06:53 AM. Reason: added stuff :)
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