I started with the notion that if $n$ and $n+2$ are prime, then: \begin{equation} \begin{cases} n\mod{2}=1 \\ (n+2)\mod{2}=1 \end{cases} \end{equation} From this I deduced that $(n+1)\mod{2}=0$. So I have that $2|(n+1)$. How to show that $3|(n+1)$?
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Hint :
Observe that, modulo $6$, a prime number greater than $3$ can only be congruent to $1$ or $5$. If both $n$ and $n+2$ are (twin) primes, what can you deduce?
Bernard
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@PM2Ring: Right. I forgot to add this, as the O.P. already did. I've fixed it. Thanks! – Bernard Apr 01 '21 at 10:00
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Oh, I see! For $n$ I have to options: $n=6k+1$ and $n=6k+5$. If we go with the first one, then $n+2=6k+3$ thus not prime. So from the second option, we get $n+1=6k$. Thank you! – mcas Apr 01 '21 at 10:38
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If $n>3$, then
$ n \equiv 0 \bmod 6 \implies $ $n$ not prime
$ n \equiv 1 \bmod 6 \implies $ $n+2$ not prime
$ n \equiv 2 \bmod 6 \implies $ $n$ not prime
$ n \equiv 3 \bmod 6 \implies $ $n$ not prime
$ n \equiv 4 \bmod 6 \implies $ $n$ not prime
$ n \equiv 5 \bmod 6 \implies $ $n+1$ divisible by $6$
lhf
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Now, you have ${n,n+1,n+2}$ and you know that both $n$ and $n+2$ are primes (therefore not multiples of $3$). Where is the multiple of $3$?
– user3733558 Apr 01 '21 at 10:20