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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 !

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  • $\begingroup$ Maybe it was meant to consider $z^{-1}=-x^{-1}-y^{-1}$,though this is just a guess. $\endgroup$
    – kingW3
    Mar 25, 2017 at 15:24
  • $\begingroup$ Note that the algebra you did doesn't reveal anything new, since "the inverse of $x$ is $z^{-1}-y^{-1}$" follows immediately from your supposition. $\endgroup$ Mar 25, 2017 at 15:26
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    $\begingroup$ You seem to have left out the assumption that $m$ is odd. If $m$ is even we have elements of order three, and the claim is false. If $\omega$ is of order three and $\{x,y,z\}=\{1,\omega,\omega^2\}$ then $x+y+z=0$ as well as $x^{-1}+y^{-1}+z^{-1}=0$. OTOH, if $m$ is odd, then $\omega\notin\Bbb{F}_{2^m}$, and Misha's argument works. $\endgroup$ Mar 25, 2017 at 17:14
  • $\begingroup$ @JyrkiLahtonen Indeed !! $\endgroup$
    – Maman
    Mar 25, 2017 at 20:50

2 Answers 2

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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.

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  • $\begingroup$ Nice where does the idea of multiplying by $w+1$ come ? $\endgroup$
    – Maman
    Mar 25, 2017 at 20:51
  • $\begingroup$ You might recognize the more-standard factorization $w^3-1 = (w-1)(w^2+w+1)$ (over fields of any characteristic), which I've changed to $w+1$ to avoid ever writing $-$. (It's a special case of $w^k-1 = (w-1)(w^{k-1}+w^{k-2}+\dots+w+1)$.) $\endgroup$ Mar 25, 2017 at 20:55
  • $\begingroup$ It was so technical to think about that key idea $\endgroup$
    – Maman
    Mar 25, 2017 at 22:31
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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 $2\mid m$).

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