reducing $x^4+1$ in $\mathbb{Z}_p[x]$ 
Possible Duplicate:
reducible polynomial modulo every prime 

Ok so we have to prove the following: If $R=\mathbb{Z}_p$ for $p$ a prime, then, $x^4+1$ is reducible over $R[x]$. This is are my ideas (please let me know if there is an easier way to tackle this problem):
First if $p=2$ then we know that $x+1$ is a factor since $1^4+1=0$ and it follows that $x-1=x+1$ is a factor. 
Hence from now on let us assume $p$ is odd. First if $x^4+1=(x^2+ax+b)(x^2+cx+d)$, I have that the leading coefficients in the factors is $1$ since I can always force this, and now I obtain the following:
$bd=1$
$ad+bc=0$
$d+ac+b=0$
$a+c=0$ 
Where the above are obtained by multiplying out the factors and setting the coefficients for the value they ought to be. 
However, we can use the last equation and plug in the second equation, to obtain $a(b-d)=0$ and since $\mathbb{Z}_p$ is a field, then we know it has no zero divisors. 
If $a=0$ then we have that $b=-d$ and hence $-d^2=1$, which means that $d^2=-1$. By using quadratic residues (here is where I worry since my class has not cover that) we have that if $p\equiv 1\pmod{4}$ then we know that (The Legendre Symbol) $(\frac{a}{p})=\bar{a}^{(p-1)/2}$, using $a=-1$, we have that $(\frac{a}{p})=1$, so $-1$ is not a quadratic residue, i.e., it is a square., hence
$x^4+1=(x^4-(-1))=(x^2-d)(x^2+d)$
I am stuck in the case when $p\equiv 3\pmod{4}$. Any help, comments, hints, or thoughts would be appreciated!  

If $p \equiv 1 \pmod{4}$, we have the factorization $$x^4 + 1 = (x^2 + i)(x^2 - i),$$ where $i \in \mathbb{F}_p$ is the element such that $i^2 = -1$.

When $p \equiv 3 \pmod{4}$, if $2$ is a square $\bmod{p}$ (say $a^2 = 2$), then 
$$
x^4 + 1 = (x^2 + ax + 1)(x^2 - ax + 1).
$$
If $2$ is not a square, then $-2$ is a square $\bmod{p}$ (see my answer here for a proof).  If, say $b^2 = -2$, then
$$
x^4 + 1 = (x^2 +bx -1)(x^2 -bx -1).
$$
