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Prove using contradiction that any prime number greater than $3$ is of the form $6n \pm 1$.

Thanks for any help

user141592
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Christie
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2 Answers2

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Can you divide $(6n+2)$ by anything? What about $(6n+3)$?

turkeyhundt
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  • We usually do proof by contradiction questions using negations and disproving this, is there any way to do it like that? – Christie Dec 17 '14 at 20:16
  • @Christie Suppose we have a prime that is not of the form $6n \pm 1$. Then $p$ is of the form $6n + r$ for $r = 0,2,3,4$ and... – Arthur Dec 17 '14 at 20:19
  • I don't know how to do that formally. Maybe somebody else on here will. You do know that every number is one of the following:$(6n\pm1),(6n\pm2),(6n\pm3)$. Let's say $(6n\pm2)$ is prime, but we know it's not because it's divisible by 2. So let's say $(6n\pm3)$ is prime. Again, we know it is not because it's divisible by 3. So all primes must be $(6n\pm1)$. Oh, also rule out $(6n+0)$. – turkeyhundt Dec 17 '14 at 20:22
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Assume that $N$ is a prime larger than $3$ and not of the form $6n\pm1$:

  • $N\equiv0\pmod6\implies N$ is divisible by $6\implies N$ is not a prime
  • $N\equiv2\pmod6\implies N$ is divisible by $2\implies N$ is not a prime (or $N=2$)
  • $N\equiv3\pmod6\implies N$ is divisible by $3\implies N$ is not a prime (or $N=3$)
  • $N\equiv4\pmod6\implies N$ is divisible by $2$ and $N\geq4\implies N$ is not a prime

Therefore, $N$ is a prime larger than $3\implies N\equiv1,5\pmod6\implies N\equiv\pm1\pmod6$

barak manos
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