Find all integers n > 1 so that n, n + 2, n + 4 are all prime.
Have no idea what to do, anyone know how to solve this one?
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Hint : One of the three numbers is divisible by $3$.
Peter
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Why must one of the numbers be divisible by 3? Can you explain, thx. – Leon K Oct 07 '15 at 22:37
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If $n$ is divisible by $3$ , we are done. If the remainder after the division by $3$ is $1$, the number $n+2$ is divisible by $3$. If the remainder is $2$, the number $n+4$ is divisible by $3$. – Peter Oct 07 '15 at 22:41
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1The three numbers must be primes! The only case for this is $(3,5,7) – Piquito Oct 07 '15 at 23:00
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The two first must be twin numbers. For the rest I had a lapse. Sorry.
The question is then Find all the triples of consecutive primes such that the two first ones and the two last ones are both twin numbers.
The integer $n$ can be equal to $3m, 3m+1, 3m+2$. Taking these three only possibilities we see clearly that of $n,n+2,n+4$ one of them is necessarily divisible by 3.Hence the only solution $(3,5,7)$. This is a very easy problem indeed.
Piquito
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I think you've misread the question; your remarks seem to be about finding $n$ such that $(n, n+2, n+6)$ are all prime – Gregory J. Puleo Oct 07 '15 at 22:48
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