If $p$ and $q$ are prime numbers larger than $2$, then $pq + 1 $ is never prime I am trying to prove the following: If $p$ and $q$ are prime numbers larger than $2$, then $pq + 1 $ is never prime. Any ideas?
 A: Curious that none of the answers so far gets this very simple argument completely right (without implicitly assuming at some point that primes are always odd). [At least they didn't at the time of writing.]
Let $p,q$ be odd primes. Then since $p,q$ are odd, $pq+1$ is even. If it were prime, one would have $pq+1=2$ since that is the only even prime number. But then $pq=1$ while odd primes are${}\geq3$, clearly a contradiction.
(Alternatively for the final phrase: "but $1$ has no prime divisors at all" or "but $1$ is not a prime number".)
A: If $p$ and $q$ are primes larger than $2$, then they are odd. It follows that $pq+1$ is even and larger than $2$, so $pq+1$ is composite.
A: Another kind of hint: If you have no idea where to start in this kind of problems, you can always try some small values and try if you can see a pattern.
$$
\begin{eqnarray*}
3 \cdot 3 + 1 &=& 10 \\
3 \cdot 5 + 1 &=& 16 \\
5 \cdot 5 + 1 &=& 26 \\
3 \cdot 7 + 1 &=& 22 \\ 
5 \cdot 7 + 1 &=& 36 \\
&\vdots&
\end{eqnarray*}
$$
A: Hint: 
Since $p,q > 2$ and the product of two odd numbers is odd then $pq + 1$ must be even.
A: If $p,q>2$ and $p,q\in\Bbb P$ then $2\not\mid p,q$. This is obvious.
If $p,q$ are odd then $2\mid (p*q+1)$, what make obvious that is not prime.
