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I'm having a lot of difficulties with this proof. Can someone please solve it and explain to me what's going on at each step? Thank you!

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marked as duplicate by Martin Sleziak, Namaste discrete-mathematics Apr 27 '17 at 1:31

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    $\begingroup$ Hint: $n, n+2, n+4$ all have different remainders when you divide by $3$. $\endgroup$ – vadim123 Apr 13 '15 at 22:06
  • $\begingroup$ I proved the theorem for n = 2. Is that not enough? $\endgroup$ – Adam Apr 13 '15 at 22:09
  • $\begingroup$ @Adam - no... for example, why doesn't 37 work? or 59? or 107? $\endgroup$ – Joffan Apr 13 '15 at 22:14
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Following my hint in the comments, one of the numbers $\{n,n+2, n+4\}$ must always be a multiple of $3$. However the hypotheses of the problem are that all three are prime. Hence one of them must be the specific prime $3$, as that is the only prime number that is also a multiple of $3$. So there are three cases:

  1. $n=3$. Excluded, since $p>3$ forbids $p=3$.

  2. $n+2=3$. Then $n=1$, which is not prime.

  3. $n+4=3$. Then $n=-1$, which is also not prime.

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  • $\begingroup$ +1 but why did you use "n" at all? $\endgroup$ – Kim Jong Un Apr 13 '15 at 22:19
  • $\begingroup$ Because $p$ is always prime, while $n$ may or may not be. $\endgroup$ – vadim123 Apr 13 '15 at 22:51
  • $\begingroup$ @vadim123 why must one of the numbers be a multiple of 3? $\endgroup$ – shoestringfries Oct 13 '15 at 1:18
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    $\begingroup$ @shoestringfries, because $n, n+2, n+4$ all give different remainders when you divide by $3$, and there are only three possible remainders. $\endgroup$ – vadim123 Oct 13 '15 at 1:30
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$p$ must be odd and greater than $3$. Thus

$$~~~~~p + 2 \equiv 2 \pm 1 \pmod 3$$ $$\Leftrightarrow p + 4 \equiv 1 \pm 1 \pmod 3$$

If the sign is plus, $p+2 \equiv 3 \equiv 0 \pmod 3$. If the sign is minus, $p+4 \equiv 0 \pmod 3$. In either case, one of these numbers is divisible by $3$, and since they can't be $3$ themselves by assumption, they cannot be prime.

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  • $\begingroup$ So the theorem is disproven? $\endgroup$ – Adam Apr 13 '15 at 22:21
  • $\begingroup$ @Adam Yes, through reductio ad absurdum. $\endgroup$ – Columbo Apr 13 '15 at 22:22
  • $\begingroup$ Got it. Thanks mate! $\endgroup$ – Adam Apr 13 '15 at 22:36

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