>Show that if $n$ is odd then $2$ is not a square in $GF(5^n)$

Question

Show that if $n$ is odd then $2$ is not a square in $GF(5^n)$

MY TRY: I tried induction on $n$.

I took $n=1$.

Then I have the finite field $\Bbb Z_5=\{0,1,2,3,4\}$

Now $1^2=1=4^2$ and $2^2=4,3^2=4$ so $2$ is not a perfect square here.

But I don't understand how to proceed by induction?

Can I get some help.

Answer 1

$2$ isn't a square in $GF(5)$, so that adjoining a square root to $GF(5)$ gives a quadratic extension $GF(25)$. If $2$ has a square root in $GF(5^n)$ then $GF(25)$ must be a subfield of $GF(5^n)$ and so $5^n$ is a power of $25$.