Note that the Galois group of $f(x) = x^4+1$ acts as the Klein four-group consisting of the identity and double transpositions on its roots. It follows that for any prime $p$, the polynomial $f(x)$ splits completely or into a product of two quadratic factors.

Straightforward by quadratic residues. We first prove some well-known results of quadratic residues: Lemma. For prime number $p \equiv 1 \pmod{4}$, there exists an integer $x$, such that $x^2 \equiv -1 \pmod{p}$. Proof. We're enough to prove $\left( \frac{-1}{p} \right)=1$. By Euler's criterion, we have $$\left( \frac{-1}{p} \right) = (-1)^{\frac{p-1}{2}} \equiv \begin{cases} 1 & p \equiv 1 \pmod{4} \\ -1 & p \equiv 3 \pmod{4} \end{cases}$$Hence, proved. $\blacksquare$ Lemma. For prime number $p \equiv 7 \pmod{8}$, there exists an integer $x$, such that $x^2 \equiv 2 \pmod{p}$. Proof. We're enough to prove $\left( \frac{2}{p} \right)=1$. By Gauss's lemma, we have $\left( \frac{2}{p} \right) = (-1)^n$ where $n$ is number of integers greater than $\frac{p}{2}$ in the set $\{2, 4, 6, \ldots, p-1\}$. We can easily see that $n=\frac{p-1}{2}-\left[ \frac{p}{4} \right]$, and we get $$\left( \frac{2}{p} \right) = (-1)^n \equiv \begin{cases} 1 & p \equiv \pm 1 \pmod{8} \\ -1 & p \equiv \pm 3 \pmod{8} \end{cases}$$Hence, proved. $\blacksquare$ Lemma. For prime number $p \equiv 3 \pmod{8}$, there exists an integer $x$, such that $x^2 \equiv -2 \pmod{p}$. Proof. We're enough to prove $\left( \frac{-2}{p} \right)=1$. As we proved earlier, observe that $\left( \frac{-1}{p} \right) = -1$ and $\left( \frac{2}{p} \right) = -1$. Then, we get $$\left( \frac{-2}{p} \right) = \left( \frac{-1}{p} \right) \left( \frac{2}{p} \right) =1$$as desired. $\blacksquare$ Now, back to the problem and expand the given equation to get $$n^4+1-(n^2+an+b)(n^2+cn+d)=-(a+c)n^3-(ac+b+d)n^2-(ad+bc)n-(bd-1).$$Then, we are enough to prove that there exists integers $a, b, c, d$ such that $$a+c \equiv ac+b+d \equiv ad+bc \equiv bd-1 \equiv 0 \pmod{p}.$$Letting $c \equiv -a \pmod{p}$, we get $$a^2 \equiv b+d \pmod{p}, \quad a(d-b) \equiv 0 \pmod{p}, \quad bd \equiv 1 \pmod{p}.$$Divide it into the following four cases: Case 1. If $p=2$, take $a=c=0, b=d=1$ then it satisfies the above condition. Case 2. If $p \equiv 1 \pmod{4}$, there exists integer $t$ such that $t^2 \equiv -1 \pmod{p}$. Take $a=c=0, b=t, d=-t$ then it satisfies the above condition. Case 3. If $p \equiv 3 \pmod{8}$, there exists integer $t$ such that $t^2 \equiv -2 \pmod{p}$. Take $a=t, c=-t, b=d=-1$ then it satisfies the above condition. Case 4. If $p \equiv 7 \pmod{8}$, there exists integer $t$ such that $t^2 \equiv 2 \pmod{p}$. Take $a=t, c=-t, b=d=1$ then it satisfies the above condition. Summing up all those cases, for all prime number $p$, there exist four integers $a, b, c, d$ that satisfies the given condition.

Consider the polynomial $x^4+1\in\mathbb{F}_p$, $F$ its splitting field and $G$ its Galois group.As $x^4+1|x^8-1|x^{p^2}-1\Rightarrow F\in\{\mathbb{F}_p,\mathbb{F}_{p^2}\}$,and since $|G|=[F:\mathbb{F}_p]\Rightarrow |G|\in\{1,2\}$.We know that $G$ acts as a cyclic group over $f$'s roots of an ireducible factor,so the degree of those must be at most $2$.This solves the problem.

Kind of folklore. At least one of $2$, $-1$ and $-2$ is a quadratic residue since their product $4$ is such. If $s^2 \equiv 2 \pmod p$, then $x^4 + 1 \equiv (x^2 + 1)^2 - 2x^2 \equiv (x^2+1) - (sx)^2 = (x^2 - sx+1)(x^2+sx+1)$. If $s^2 \equiv -2 \pmod p$, then $x^4 + 1 \equiv (x^2 - 1)^2 + 2x^2 \equiv (x^2 - sx - 1)(x^2 + sx - 1)$. Finally, if $s^2 \equiv -1 \pmod p$, then $x^4 + 1 \equiv (x^2 - s)(x^2 + s)$.

\[p|n^3(a+c)+n^2(b+d+ac)+n(ad+bc)+bd-1\] $i)p \equiv 1 (mod 4)$ Choose $(a,b,c,d) \equiv (0,b,0,-b)$ where $b^2 \equiv -1 (mod p)$ It holds because \[p|n^3.0+n^2.0+n.0-b^2-1\equiv 0\]$ii)p \equiv 3(mod 4)$ We have \[(\frac{-2}{p})=(\frac{-1}{p})(\frac{2}{p})=-(\frac{2}{p})\]which means there exists $a$ for $a^2 \equiv 2(mod p)$ or $a^2 \equiv -2(mod p)$ If $a^2 \equiv -2 (mod p)$ exists, then choose $(a,b,c,d)\equiv (a,-1,-a,-1)$ where $a^2 \equiv -2$ It holds because \[p|n^3p+n^2(-2+ap-a^2)+n(-a+a)+1-1 \equiv 0\] If $a^2 \equiv 2 (mod p)$ exists, then choose $(a,b,c,d)\equiv (a,1,-a,1)$ where $a^2 \equiv 2(mod p)$ It holds because \[p|n^3p+n^2(ap-a^2+2)+n(a+p-a)+1-1 \equiv 0\] $iii)p=2$ Choose $(a,b,c,d)=(1,1,1,1)$

Good question for using quadratic residues. Not appropriate for #5, but I think it is a good problem.