Find all positive integers $n$ such that $n$,$n + 2$, and $n + 4$ are all primes... Find all positive integers $n$ such that $n$,$n + 2$, and $n + 4$ are all primes.


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*Having a tough time with this problem, I feel that brute force is a possibility especially considering that my professor said that there are only a small amount of numbers that fit this category, but there has to be a better way. Any help or suggestions are greatly appreciated. I did see a similar post to this question but it did not help me solve this nor did it look to find the same thing, please do not mark this as a duplicate. 

 A: Note that one of the three numbers must be a multiple of $3$. So the only possible choice is that one of them is $3$ itself. So $3,5,7$
A: Brute force can nudge you towards the answer, but you still have to look at and think about the results of brute force. Have your computer give you a few triples factorized and analyze the results.
Obviously, $n$ has to be a prime number. So there is no need to test $n = 0$ or $n = 1$.
Then, with $n = 2$, we see that $n + 2$ can't prime because that's obviously divisible by 2. No need to look at $n = 4$ either. In fact, we don't need to consider any other even $n$.
Moving on to $n = 3$, we obtain the primes 3, 5, 7. Ding, ding, ding!
With $n = 5$, we get the prime 7 but also $9 = 3^2$.
And with $n = 7$, we get $n + 4 = 11$, but $n + 2 = 9 = 3^2$.
Look at this problem modulo 6: if $n$ is odd, it has to be 1 or $5 \pmod 6$ to not be a multiple of 3. But if $n \equiv 1 \pmod 6$, then $n + 2 \equiv 3 \pmod 6$, or if $n \equiv 5 \pmod 6$, then $n + 4 \equiv 3 \pmod 6$.
So this problem has exactly one solution, or exactly two if negative integers are allowed.
