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?
If $a>2$, then $a-1,a+1>1$, and therefore $(a-1)(a+1)$ cannot be prime (it has two factors greater than $1$). And if $a=1$, then $(a-1)(a+1)=0$, which is not prime. But if $a=2$, then $(a-1)(a+1)=3$, which is prime. Therefore, the answer is $\{2\}$.
Notice that $$(a+1)(a-1)$$ can only be prime if exactly one of the factors is $1$.
Let $a+1=1$. Then $$a=0\implies (a+1)(a-1)=1(0-1)=-1$$ which cannot be possible.
Now let $a-1=1$. Then $$a=2\implies (a+1)(a-1)=(2+1)(1)=3$$ which is prime.
Hence $a=2$ is the only solution.
Assuming $a\geq0$, $(a+1)(a-1)$ is not prime if both $(a+1)$ and $(a-1)$ are greater than $1$. So we must have $(a-1)\leq1$, which means $a\leq2$. That gives the only solution $a=2$, for if $a=1$ then $a-1=0$.
If you allow non integer values of $a$ then for any prime $p$ we have that when $a=\sqrt{p+1}$ that $(a-1)(a+1)=p$ so the answer is for infinitely many values of $a$.
If you assume that $a$ is an integer then either $a-1=1$ and $a=2$ or $a-1=-1$ but then $a=0$ and $(a-1)(a+1)=-1$ not a prime number or $a+1=1$ but then $a=0$ again and if $a+1=-1$ then $a=-2$ which gives $(a-1)(a+1)=3$ then the answer is two values $\{2,-2\}$.
If you assume that $a$ is a positive integer then the solution is one value $\{2\}$.
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.
