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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3Hint: $n, n+2, n+4$ all have different remainders when you divide by $3$. – vadim123 Apr 13 '15 at 22:06
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I proved the theorem for n = 2. Is that not enough? – Adam Apr 13 '15 at 22:09
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@Adam - no... for example, why doesn't 37 work? or 59? or 107? – 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:
$n=3$. Excluded, since $p>3$ forbids $p=3$.
$n+2=3$. Then $n=1$, which is not prime.
$n+4=3$. Then $n=-1$, which is also not prime.
vadim123
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1@shoestringfries, because $n, n+2, n+4$ all give different remainders when you divide by $3$, and there are only three possible remainders. – vadim123 Oct 13 '15 at 01:30
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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.
Columbo
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