For how many positive integers $a$ is $a^4−3a^2+9$ a prime number? I understand that there are many posts on the problems similar to mine. I have tried my best, but still get different answers from the answer sheet. Can anyone help me? Also is there a simple way to find $a$?

For how many positive integers $a$ is $a^4-3a^2+9$ a prime number?

Here is what I did: $$a^4-3a^2+9=(a^2+3+3a)(a^2+3-3a)$$ To find $a$, I looked at the following situations:
$$a^4-3a^2+9=1$$ 
$$a^4-3a^2+9=3$$ 
$$a^4-3a^2+9=5$$
$$a^4-3a^2+9=7$$
$$a^4-3a^2+9=11$$
$$a^4-3a^2+9=13$$
$$....$$
 A: Here is a fundamental fact about prime numbers: IF: a prime number $p$ factors as $p = xy$, where $x$ and $y$ are integers, THEN: $x = \pm 1$ or $y = \pm 1$.
You have already correctly noticed that
$$
a^4-3a^2+9=(a^2+3+3a)(a^2+3-3a).
$$
Assuming that $a^4 - 3a^2 + 9$ is prime, it follows by the above result that there are two possibilities:
\begin{align*}
a^2 + 3 + 3a &= \pm 1 \\
a^2 + 3 - 3a &= \pm 1.
\end{align*}
This is really four possibilities in disguise:
\begin{align*}
a^2 + 3 + 3a &= 1 \\
a^2 + 3 + 3a &= -1 \\
a^2 + 3 - 3a &= 1 \\
a^2 + 3 - 3a &= -1.
\end{align*}
Now, your job is to solve for $a$ in each of these four cases. Then, check for each case whether the $a^4 - 3a^2 + 9$ is actually prime.
A: HINTS
Your factorisation is the key: $a^4 - 3a^2 + 9 \equiv (a^2-3a+3)(a^2+3a+3)$.
Since $a$ is an integer, so are both $a^2-3a+3$ and $a^2+3a+3$.
If both $a^2-3a+3$ and $a^2+3a+3$ are bigger than $1$ then $a^4 - 3a^2 + 9$ will have two positive integer factors larger than one, i.e. $a^4 - 3a^2 + 9$ won't be prime. (Consider only positive $a$.)
You could prove this by exhaustion. Look at when $a^2-3a+3 > 1$ and when $a^2+3a+3>1$.
A: You already have the product $p = \left(a^2-3 a+3\right) \left(a^2+3 a+3\right)$, and since for both factors you have $0 < a^2-3 a+3 < a^2+3 a+3$ for positive $a$ (check), and since a prime number is only divisible by $1$ and itself, you have $a^2-3 a+3 = 1$ and $a^2+3 a+3 = p$. Solving first equation gives you $a=1$ and $a=2$ and these in turn give you two primes $p=7$ and $p=13$.
A: One must have $$a^2+3a+3=\pm1\\a^2-3a+3=\pm1$$ From these we get, discarding the constant term $4$, only two equations
$$a^2+3a+2=0=(a+1)(a+2)\\a^2-3a+2=0=(a-1)(a-2)$$ which give
both the primes $7$ and $13$.
