TITLE: For how many values of $a$ will $(a + 1)(a - 1)$ be prime?
QUESTION [5 upvotes]: For how many values of $a$ will the number $(a+1)(a-1)$ be prime?
I showed that $a$ must be even because if it is odd then $(a-1)(a+1)$ will be even and therefore not prime. If it is even, then I also showed that $a$ must be a multiple of $3$ because otherwise $(a-1)(a+1)$ will be a product of $3$. Seems like it's not the right method to find the number of $a$ that will give $(a+1)(a-1)$ prime?

REPLY [3 votes]: Try swapping $a$ for $n$. Then it's clearer that $n$ is an integer, for humans and computers alike. Put (n - 1)(n + 1) into Wolfram Alpha and it will give you the alternate form $n^2 - 1$.
If $n$ is odd, then $n^2 - 1$ is not only even but doubly even (divisible by $4$). And if $n$ is not divisible by $3$, we will find that $n^2 - 1$ is, leading us to find that if $n = \pm 2$, then $n^2 - 1 = 3$; otherwise $n^2 - 1$ will be divisible by at least one prime other than $3$.
So that leaves even numbers that are divisible by $3$. Since $|n| \geq 6$, we find that $n - 1$ and $n + 1$ are distinct non-unit numbers.
Well, that was the scenic route. The shorter route would have been to make the observation about $n - 1$ and $n + 1$ right off the bat.