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:
$$a^4-3a^2+9=1$$ 
$$a^4-3a^2+9=3$$ 
$$a^4-3a^2+9=5$$
$$a^4-3a^2+9=7$$
$$a^4-3a^2+9=11$$
$$a^4-3a^2+9=13$$
$$....$$

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 $p=13$.
