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If $F$ is a finite field such that every element is a square, why must $char(F)=2$?

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If every element is a square, then the map $x\mapsto x^2$ is a surjection from $F$ to itself. Since $F$ is a finite field, the map is a bijection, which means that since $-1$ and $1$ have the same image, then $-1=1$, hence the characteristic is $2$.

In fact, the converse holds as well. Suppose $F$ is a finite field of characteristic $2$: then it has $2^n$ elements, for some $n$, hence the multiplicative group of nonzero elements is $F$ of order $2^n-1$, which is odd. In particular, there is no element of order $2$ by Lagrange's theorem. The map $f\colon F-\{0\}\to F-\{0\}$ given by $f(x)=x^2$ is a group homomorphism whose kernel consists of all elements of exponent $2$; by the comment just made, the kernel is trivial, hence $f$ is one-to-one and hence onto, so every nonzero element of $F$ is a square. And of course, $0$ is a square.

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Another way: all elts of a finite field of order $\, 2n\!+\!1\,$ are squares $\Rightarrow x^{n}\! - 1\ $ has $2n$ roots (all elts $\ne 0$), contra a polynomial over a field has no more roots than its degree.

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  • $\begingroup$ @Downvoter I added more details. Please ask if something is not clear rather than downvote. $\endgroup$ – Bill Dubuque Dec 8 '19 at 1:28

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