All I've been able to deduce is that I should suppose $1+1=0$ in $F$ and then $a+a=0$ for all $a$ in $F$, i.e each element in our field is it's own inverse. And somehow, this is supposed to be impossible because there are an odd number of elements.
I'd like some help with the last part, thanks.
We've done pretty much nothing besides the axioms defining a field, and some other basic things (no zero divisors, cancellation etc)
Edit : My instructor told me (as a hint) to consider the the map $f : a -> a^2$, and then derive a contradiction using the fact that $|F| = odd$. I still don't see a way to get the contradiction.. One (perhaps) useful thing I've noted is that $f$ is a bijection, so we essentially get a copy of the same field with the map.
Let $n$ be the number of elements in the field. Let $S$ be the sum of all elements in the field.
The map $x \mapsto x+1$ is a bijection. (The inverse map is $x \mapsto x-1$.)
Then $S=\sum_x x = \sum_x (x+1) = S+n\cdot1$ and so $n\cdot 1 = 0$.
Write $n = 2q+r$ with $r \in \{0,1\}$. Then $$ 0 = n\cdot 1 = 2q \cdot 1 + r \cdot 1 = r \cdot 1 $$ If $r=1$, then $1=0$ and $n=1$. Otherwise, $r=0$ and $n$ is even.