Solution

The consecutive odd numbers 3, 5, and 7 are all primes. Are there infinitely many such "prime triplets"? That is to say, are there infinitely many prime numbers $p$ so that $p+2$ and $p+4$ are also primes? Prove your conclusion.


We argue by cases:

Consider the possible remainders produced when $p$ is divided by 3. The only possibilities are 0, 1, and 2. Consider each case seperately: $$\begin{align} p \textrm{ has remainder } 0 \textrm{ when divided by 3 }&\Rightarrow 3 \mid p\\ & \\p \textrm{ has remainder } 1 \textrm{ when divided by 3 }&\Rightarrow p = 3k+1\\ &\Rightarrow p+2 = 3k+3\\ &\Rightarrow p+2 = 3(k+1)\\ &\Rightarrow 3 \mid p+2\\ & \\p \textrm{ has remainder } 2 \textrm{ when divided by 3 }&\Rightarrow p = 3k+2\\ &\Rightarrow p+4 = 3k+6\\ &\Rightarrow p+4 = 3(k+2)\\ &\Rightarrow 3 \mid p+4\end{align}$$ Thus (with the exception of 3, 5, and 7) three consecutive odd numbers are never all prime. One of them must always be divisible by 3.

QED.