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Im practicing for an exam, can anyone help me with this?

Let $p$ be prime and let $x=y^p-y$, for $y \in \mathbb{F}_{p^2}$. Prove: $x^2 \in \mathbb{F}_p$

user132531
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    $(y^p-y)^p=-(y^p-y)$ – user8268 Jun 04 '14 at 18:23
  • @user8268: You could consider posting a (possible slightly more verbose) version of your hint as an answer :-). To the OP: can you take advantage of that hint? Do you know of polynomial equation such that all its roots ar in the prime field? Does $x^2$ satisfy that equation? – Jyrki Lahtonen Jun 04 '14 at 21:06
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    @JyrkiLahtonen Perhaps easier than thinking about a polynomial whose roots are in the prime field, one could use user8268's hint to say that the product of $x$ and its conjugate $x^p = -x$, the additive inverse of the "constant" term in the minimal polynomial of $x$ in $\mathbb F_p[z]$ must be a member of $\mathbb F_p$. – Dilip Sarwate Jun 05 '14 at 00:49
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    $(y^p-y)^p=y^{p^2}-y^p=-(y^p-y)$ \

    $((y^p-y)^p)^2=(-(y^p-y))^2$\

    $((y^p-y)^2)^p=(y^p-y)^2$\

    All the solutions of $z^p=z$ are in $\mathbb{F}_p$, so $x^2=(y^p-y)^2$ is as well.

     – user132531 Jun 05 '14 at 02:54
  • Correct, @user132531. You, too, could post it as an answer (may be there is a time window when the original poster cannot? but that will expire soon). – Jyrki Lahtonen Jun 05 '14 at 04:27

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Since no one else seems inclined to answer and @user8268's comment is the key to the solution, here is my own comment converted into an answer. I have marked it as Community wiki.

Since $x = y^p-y \in \mathbb F_{p^2}$ has one conjugate $x^p = (y^p-y)^p = y^{p^2}-y^p = y - y^p = -x$, the minimal polynomial of $x$ in $\mathbb F_p[z]$ is $(z-x)(z-(-x)) = (z-x)(z+x) = z^2-x^2$ which shows that $-x^2$, and hence $x^2$, belongs to $\mathbb F_p$.

Dilip Sarwate
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