Is the following statement true ?
If so, please explain why.
If K is an algebraically closed field, then K is an infinite set.
Every finite field has order $\displaystyle p^n$ for some n. The underlying multiplicative group is always cyclic (of order $\displaystyle p^{n}-1$). Thus, $\displaystyle g^{p^n-1}=1$ for all $\displaystyle g \in \mathbb{F}_{n}\setminus \{0\}$.
Therefore, if $\displaystyle p \neq 2$, the equation $\displaystyle x^{p^n-1}+2$ will not have a root in $\displaystyle \mathbb{F}_n$.
If $\displaystyle p=2$ then this trick doesn't work...however, if $\displaystyle F=\{a_1, a_2, \ldots, a_{m}\}$ is your finite field, then the polynomial $\displaystyle (x-a_1)(x-a_2)\ldots (x-a_{m})+1$ has no roots in $\displaystyle F$ (this works for all p).