<s>[INST] 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:
$$a^4-3a^2+9=1$$ 
$$a^4-3a^2+9=3$$ 
$$a^4-3a^2+9=5$$
$$a^4-3a^2+9=7$$
$$a^4-3a^2+9=11$$
$$a^4-3a^2+9=13$$
$$....$$
 [/INST] 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 $a^2+3a+3>1$.
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