I came across a problem which had a line of explanation as follows :
Let $a \in GF(p). a$ is a non square in $GF(p),~ p \neq 2 \implies \nexists~ b\in GF(p)~~|~~a =b^2 $.
But, is it really possible that $GF(p)$ contains a non square element? The non zero elements of $GF(p)$ form a group ( cyclic ) under multiplication and hence, the mapping $x \rightarrow x^2$ is a one-one, onto mapping.
Hence, $\forall~~x \in GF(p)~~x = y^2~~|~~ y \in GF(p)$.
And $0^2$ is of course $0$ in $GF(p)$.
Hence, can there really exist non square elements in $GF(p)$?
EDIT: I realize from Andreas's comment that the mapping $x \rightarrow x^2$ is not one-one. However, isn't this an example of a Frobenius mapping .
$\Psi:GF(p) \rightarrow GF(p)$ s.t. $a \rightarrow a^p$ where $p$ is a prime is considered a ring automorphism. Is there a contradiction?
Thank you for your help..