How to prove that $x^{-1}+y^{-1}+z^{-1} \ne 0$? Let $m$ odd integer, $m\ge 3$ and $x,y,z\in \mathbb{F}_{2^m}^*$ such that $x+y+z=0$. I have to prove that $x^{-1}+y^{-1}+z^{-1} \ne 0$.
Hint : by absurd suppose that $z^{-1}=x^{-1}+y^{-1}$ and consider the element $xy^{-1}$.
So multiplying by $x$ I obtain : $xz^{-1}=1+xy^{-1} \Leftrightarrow x(z^{-1}-y^{-1})=1$. That would mean that the inverse of $x$ is $(z^{-1}-y^{-1})$. But I don't know how to conclude.
Moreover I don't know if it helps but there are no elements of order $3$.
Thanks in advance !
 A: Here's what's up with $w = xy^{-1}$.
Suppose for the sake of contradiction that $x+y+z=0$ and $x^{-1} + y^{-1} + z^{-1} =0$, or $z = x+y$ and $z^{-1} = x^{-1}+y^{-1}$. Multiplying these together, we get
$$1 = zz^{-1} = (x + y)(x^{-1} + y^{-1}) = 1 + x^{-1}y + xy^{-1} + 1 = x^{-1}y + xy^{-1}.$$
If we let $w = xy^{-1}$, this means $w + w^{-1} = 1$, which we can rearrange to get $w^2 + w + 1 = 0$. Multiplying by $w+1$, we have $w^3 + 1 = 0$, or $w^3 = 1$.
But we're given that there are no elements of order $3$, contradiction.
A: As an alternative you can think in terms of locator polynomials. In this case consider
$$
f(T)=(T-x)(T-y)(T-z)=T^3-(x+y+z)T^2+(xy+yz+zx)T-xyz\in\Bbb{F}_{2^m}[T].
$$
The condition $x+y+z=0$ tells us that the quadratic term of $f(T)$ vanishes. Because
$$
\frac1x+\frac1y+\frac1z=\frac{xy+yz+zx}{xyz}
$$
the sum in the left can vanish only when the linear of $f(T)$ vanishes as well.
So we know that $f(T)$ is of the form
$$
f(T)=T^3-c
$$
with $c=xyz\neq0$. For $f(T)$ to have three solutions in $\Bbb{F}_{2^m}$ it is necessary that $c$ has three distinct cube roots in it. The ratio of two cube roots is a third root of unity, so this can only happen when $\Bbb{F}_{2^m}^*$ contains elements of order three (which is equivalent to $2\mid m$).
