$\sigma^4=\operatorname{id}$ $\sigma^3(\alpha)+\sigma(\alpha)=\sigma^2(\alpha)+\alpha$ then $\sigma^2=\operatorname{id}$ Let $\sigma$ an automorphism of a field $F$ such that $\sigma^4=\operatorname{id}$ and for all $\alpha\in F$ $$\sigma^3(\alpha)+\sigma(\alpha)=\sigma^2(\alpha)+\alpha.$$
Show that $\sigma^2=\operatorname{id}$.
I tried it many hours. I have not idea which is the trick.
 A: Artin's Lemma of linear independence of characters says that a set of distinct characters is linearly independent. The given equation manifestly states that the automorphisms $\sigma^i,i=0,1,2,3,$ are linearly dependent. Therefore they cannot be all distinct.
Thus the order of $\sigma$ is strictly less than $4$. On the other hand the order must be a factor of four. Therefore it is a factor of two.
A: Now that a nice solution has been given, let me give a direct solution,
at least when char. F $\neq 2$.
As Jim observed in the comments,
the minimal poly. of $\sigma$ divides $(X-1)(X^2 + 1)$.
Suppose now that the char. $\neq 2$.  Then $X - 1$ and $X^2 + 1$ are coprime, and so we may decompose $F$ into the direct sum of a space where $\sigma  = 1$ (the fixed field of $\sigma$) and of a space where $\sigma^2  + 1 = 0$.
Since the gcd. of $(X^2 -1)$ and $(X-1)(X^2 +1)$ equals $X-1$, we also
see that the fixed field of $\sigma^2$ equals the fixed field of $\sigma$.
Now if $\alpha$ lies in the space where $\sigma^2 + 1 = 0$, then $\sigma^2(\alpha^2) = (\sigma^2(\alpha))^2 = (-\alpha)^2 = \alpha^2,$ and so $\alpha^2$ 
is fixed by $\sigma^2$.  Thus by the preceding remark it is fixed by $\sigma$,
i.e. $\sigma(\alpha)^2 = \sigma(\alpha^2) = \alpha^2$.  But then $\sigma(\alpha)  = \pm \alpha$.  Thus $\sigma^2 - 1$ also acts by zero on the space where $\sigma^2 + 1$ acts by zero.  The only way this is possible is if this space
itself equals zero, i.e. if $F = F^{\sigma}$.

This argument is not so different to the general argument via independence of characters.  The key fact we used, beyond general linear algebra, is that 
$\sigma(\alpha^2) = \sigma(\alpha)^2$, which allows us to produce new eigenvalues from old.  More precisely, on the space where $\sigma^2 + 1 = 0,$
the eigenvalues are $\pm  i$.  So if $\alpha$ is an eigenvector for 
one of these eigenvalues, then $\alpha^2$ is an eigenvector with eigenvalue
$-1$.  But this eigenvalue isn't allowed.  Thus $\pm i$ can't appear as eigenvalues after all.   (In the actual argument above, I didn't argue with eigenvalues because $F$ may not contain $4$th roots of unity.  But this eigenvalue computation is what is behind the argument, and is closely related 
to independence of characters.)
