TITLE: Proof that multiplication of two numbers plus $2$ is a prime number
QUESTION [6 upvotes]: Let's say we have odd prime numbers $3,5,7,11,13, \dots $ in ascending order  $(p_1,p_2,p_3,\dots)$. Prove that this sentence is true or false : 
For every $i$,
$$p_i p_{i+1}+2$$
is a prime number.
Any ideas how can I prove this?

REPLY [3 votes]: Building off of Robert Soupe's answer, thinking about modular arithmetic is a great way to come up with counterexamples.
The first counterexample I found was $31, 37$ - both equal $1$ (mod $3$), so when we multiply them and add $2$, we get a multiple of $3$. I'm slow at mental multiplication, so searching for a pair of consecutive primes which were both 1 or both 2 (mod 3) was easier for me than trying some small examples.
Similarly, $p_i, p_{i+1}$ form a counterexample if:

$p_i=1$ (mod $5$), $p_{i+1}=3$ (mod $5$) - for example, $41$ and $43$.
$p_i=4$ (mod $7$), $p_{i+1}=3$ (mod $7$) - for example, $53$ and $59$ ($57$ is not prime! :P).
And so on.