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

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 $13$.
