Trying to factor $x^4-3$ over $\Bbb{F}_p$ with $p\equiv 3\mod{4}$ So I am trying to show that $f(x) = x^4-3$ will factor over $\Bbb{F}_p$ when $p\equiv 3\mod 4$, with $p\neq 3$. Specifically, I am trying to show that $f$ will factor into irreducibles such that there is at least one factor of degree 2.
My Work So Far
So if I assume that it can be factored, we would have
$$ x^4-3 = (x^2+ax+b)(x^2-ax+d) = x^4 + (b+d-a^2)x^2 + a(d-b)x + bd.$$
Because we want the linear term to be 0, we need either $a=0$ or $d-b = 0$. Now if $a=0$ we get that $b=-d$, so $-b^2 = -3$. That is, $3$ is a square. If $d-b = 0$, then we would have $b^2 = -3$, so $-3$ is a square. Because $-1$ is not a square when $p\equiv 3\mod{4}$, we know that one of $-3$ or $3$ must be a square, but not both, so $a=0$ or $d=b$, but not both.
If $a=0$, that is $3$ is a square, then we have $f(x) = (x^2 +b)(x^2-b)$. Then one of $b$ or $-b$ is a square, but not the other, so $f$ will factor into two linear terms and a quadratic. So this case works.
The trouble I am having is with the second case. If $d=b$, then we have 
$$ f(x) = (x^2 + ax + b)(x^2 -ax +b)$$ with $2b = a^2$. This is where I get stuck. How do I know that such an $a$ can exist?
Edit: If there is a better approach to solving this, that would be great to know as well. I haven't been taught any of the relevant number theory to solve this, so I'm kind of out of my element here.
 A: In your second case you need $b^2=-3$ and you know (in this case) that that is possible.  But it is possible for two different values of $b$, say $b=s$ and $b=-s$; and you know that either $2s$ or $-2s$ is a square; so choose the appropriate value of $b$ such that $2b$ is a square.
Or to put it another way: if $2$ is a square, choose the value of $b$ which is a square, then $2b$ is a square.  If $2$ is not a square then choose the value of $b$ which is not a square; then $2b=(-2)(-b)$ is a square.
A: In your second case, $3$ is not a quadratic residue mod $p$, and you want $b^2 = -3$ and $2b = a^2$.  As you say, exactly one of $3$ and $-3$ is a quadratic residue, so $-3$ is.  Thus there are two possible $b$'s, one the negative of the other.  Moreover, for exactly  one of these $2b$ will be a quadratic residue mod $p$.
A: As you pointed out in a comment the goal is equivalent to stating that the splitting field $K$ of $f(x)$ over $\Bbb{F}_p$ is $\Bbb{F}_{p^2}$.
For an alternative approach to that goal I proffer the following.


*

*If $\alpha$ (in some extension field of $\Bbb{F}_p$) is a zero of $f(x)$, the other zeros are $i^k\alpha, k=1,2,3,$ where $i$ is a square root of $-1$. Therefore $i$ is an element of the splitting field. Hence the extension degree $[K:\Bbb{F}_p]$ is at least two.

*Let $m$ be the order of $3$ in the multiplicative group $\Bbb{F}_p^*$. Lagrange's theorem tells us that $m\mid p-1$. Again, if $\alpha$ is any zero of $f(x)$ we then have $\alpha^{4m}=1$. But, $4\mid(p+1)$, so $4m\mid(p^2-1)$. The multiplicative group of $\Bbb{F}_{p^2}$ is cyclic of order $p^2-1$, so all the $4m$ zeros of $x^{4m}-1$ belong to the field $\Bbb{F}_{p^2}$. Among them $\alpha$. Hence $K\subseteq \Bbb{F}_{p^2}$ and we are done.



Obviously the same argument works for all the polynomials of the form $x^4-a$, $a\not\equiv0\pmod p$.
A: Because $p\equiv 3$ mod $p$, $-1$ is not a square mod $p$, so that $\mathbf F_p (i)$, with $i^2=-1$, is just $\mathbf F_{p^2}$, and for any $a\in \mathbf F_p ^*$,the splitting field of $X^4 - a$ is just $K=\mathbf F_p (i, \alpha)$, where $\alpha$ is a root of the polynomial. To ease the notations, write $F=\mathbf F_p, E=F(i)=\mathbf F_{p^2}$, so $K=E(\alpha)$, and let us show that $E=K$. Consider the canonical homomorphism $\phi:F^*/{F^*}^{2} \to E^*/{E^*}^{2}$. By definition, its kernel contains the class of $-1$, of order $2$. But $F^*$ is cyclic, hence $F^*/{F^*}^{2}$ has order $2$, which means that the map $\phi$ is trivial, in other words, any extension of the form $E(\sqrt a)$, with $a\in \mathbf F_p^*$, must coincide with $E$. This implies what we want.
