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So in class my professor posed a question.

Let $\mathbb F$ be a field satisfying the condition $a\to a^2$ is a permutation of $\mathbb F$. What is the characteristic of $\mathbb F.$

I was told it should be $2$, but I am not sure why. I believe the solution involved $-1$ and the something about the inverse but I am not completely sure.

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If $\text{char}(\Bbb F) \ne 2$ then $1\ne -1$, and the map $a\mapsto a^2$ cannot be injective as $-1\mapsto 1$. On the other hand if $\text{char} (\Bbb F) = 2$ then

$$|\Bbb F^\times|=2k+1$$

is odd, hence the squaring map is an automorphism of this group, in particular it is injective. As $0\mapsto 0$ this shows the map is injective on all of $\Bbb F$. It is also clearly a ring homomorphism by the binomial theorem. So $a\mapsto a^2$ is a field automorphism iff $\text{char}(\Bbb F)=2$ (as we are assuming, per the tag in the op, that we are dealing with finite fields.)

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    @MorganRodgers the question was tagged "finite fields" so I obliged accordingly. – 2017-01-31
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    @MorganRodgers in context it does, but I agree some people won't read the tag, so I'll edit accordingly. – 2017-01-31
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    A field $\mathbb{F}$ of characteristic $2$ satisfies "the map $\mathbb{F}\to\mathbb{F},\ a \mapsto a^2$ is a bijection" if and only if it is perfect. Not sure if there is a more explicit characterization. – 2017-01-31
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    @darijgrinberg the context is finite fields, see the original post's tags. In that case they're all perfect, so it doesn't say much unless one digs deeper in as it appears the op is looking for a proof of this fact. – 2017-01-31