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.