A field with an element of order $12$: $(a+a^{-1})^2=3.$ 
Let $K$ be a field and $a\in K^*$ of order $12.$ I need to prove that $(a+a^{-1})^2=3.$ 

'Progress': I expand $(a+a^{-1})^2=a^2+a^{-2}+2$ so equivalently I need to prove that $a^2+a^{-2}=1$, or the inverse of $a^2$ is $1+a^{-4}.$ I tried also Binomial theorem and working with the inverse but no success.
Not sure how can I use the fact that the order of $a$ is $12$. Any ideas?

We have $(a^6-1)(a^6+1)=0$. But $a^6\ne 1$, so $a^6+1=0$. Thus $(a^2+1)(a^4-a^2+1)=0$. Because the order of $a$ does not divide $4$, we have $a^4-a^2+1=0$. This is equivalent to what you want to show.
