Irreducible polynomials in GF(p) Given is the polynomial $\varphi(X)=X^4+1$.
Now there are two tasks:
(1) Show, that $ \varphi(X)$ is reducible in $\mathbb F_p [X]$, where $p$ is prime number with  $p \equiv1$ (mod 4).
(2) Show, that $ \varphi(X)$ is reducible in $\mathbb F_p [X]$, where $p$ is prime number with  $p \equiv3$ (mod 4).
My ideas until now:
(1) I try to find $a,b \in \mathbb F_p [X]$, such that $X^4+1 = (X^2+a)(X^2+b)$. Not to use the middleterm $+cX$ was a hint. Now $(X^2+a)(X^2+b)=X^4+aX^2+bX^2+ab=X^4+(a+b)X^2+ab$. This should equal $X^4+1$, so I know, that $a+b \equiv 0$ mod p and $ab \equiv 1$ mod p. This means, that $a$ and $b$ must be units in $\mathbb F_p$, more exactly, $a$ is the multiplicative inverse element of $b$ in $\mathbb F_p$. $a+b \equiv 0$ reasons, that $a$ is the additive inverse to $b$.
But know I don't know, how I can go on with conclusion. I hope, somebody can help me. 
 A: The two congruences give you 
$$b^2 \equiv -1 \pmod{p}$$ 
This is the same as asking when is $-1$ a quadratic residues mod $p$? If you are familiar with Legendre symbol, then perhaps you might have seen
$$\left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}=\begin{cases}1 & \text{ if } p \equiv 1 \pmod{4}\\-1 & \text{ if } p \equiv 3 \pmod{4}\end{cases}$$
So if $p \equiv 1 \pmod{4}$, then we have $a,b$ such that $a+b \equiv 0$ and $ab \equiv 1$. The existence of such $a,b$ implies that $a^2 \equiv -1 \pmod{p}$. Therefore,
$$x^4+1 =(x^2+a)(x^2-a)$$
part(2)
Claim: When $p \equiv 3 \pmod{4}$, then one of $2$ or $-2$ will be a quadratic residue. For this consider $2^{\frac{p+1}{4}}$. It can be shown that this will be the square root of either $2$ or $-2$.
If $2$ happens to be a quadratic residue mod $p$, i.e. if there exists an $a$ such that $a^2 \equiv 2 \pmod{p}$, then we can factor as follows:
$$x^4+1 = x^4+1+a^2x^2-a^2x^2=(x^2+1)^2-a^2x^2=(x^2+1+ax)(x^2+1-ax).$$
Now consider the case when $2$ is not a quadratic residue, then try to see if you can prove that $-2$ will definitely be a quadratic residue. Then you can use the idea above to complete the factorization.  
A: Alternatively, $x^{p^2}-x=x\left(x^{p^2-1}-1\right)$ is divisible by $x^8-1=\left(x^4-1\right)\left(x^4+1\right)$, as $8\mid p^2-1$ for all odd primes $p$.  As the field $\mathbb{F}_{p^2}$ with $p^2$ elements is the splitting field of $x^{p^2}-x$ and $\left[\mathbb{F}_{p^2}:\mathbb{F}_p\right]=2$, it follows that $x^4+1$ must factor into irreducible polynomials in $\mathbb{F}_p[x]$ of degree at most $2$.
