Let $(K, +, \cdot)$ a finite field so that the polynomial $P=X^2-5$ is irreducible.
Prove that:
a) $1+1 \ne 0$
b) The polynomial $P_a=X^5 + a$ is reducible $\forall a \in K$
a) This is the easy part.
Suppose $1+1=0$. Then $5=1$ therefore $P=(X-1)(X+1)$ contradiction.
b) I think the idea is to show the equation $x^5 + a=0$ has solutions $\forall a \in K$, but I failed to prove it.
Any hint for an elementary solution is appreciated.
If $a=0$,we're done. So, suppose $a\in F^\times$.
Let $p=$char$ F$. Then, $F=\Bbb{F}_{p^k}$. If $k$ is even, then, $\Bbb{F}_{p^2}\subseteq F$. However, $X^2-5$ splits over $\Bbb{F}_{p^2}$, as splitting field of a degree two polynomial is of degree two, and $\Bbb{F}_{p^2}$ is the only degree two extension of $\Bbb{F}_p$. Contradiction.
So, $k$ is odd. $x^2-5$ has no solution means $5$ is not a quadratic residue $\Bbb{F}_p$. Then, by the quadratic reciprocity law, $p$ is not a quadratic residue in $\Bbb{F}_5$. In particular $5\not\lvert p\pm 1$, so, $5\not\lvert p^k-1$. Choose a generator $g\in F^\times$. Let $-a=g^{m}$. As $5\not\lvert p^k-1$, $5$ divides one of $m,m+(p^k-1),m+2(p^k-1),m+3(p^k-1),m+4(p^k-1)$. Then, $g^\frac{m+b(p^k-1)}5$ is a root of $x^5+a$, as desired.