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Consider the following equation over the finite field $\operatorname{GF}(p^q)$ such that $r \mid p^q-1$: \begin{align} x^r=y^r \tag{1} \\ \end{align} The solutions of $(1)$ over $\operatorname{GF}(p^q)$ are:

\begin{align} x=\gamma^i\, y \quad , \quad 0\leq i \leq r-1\\ \tag{2} \end{align}

where $\gamma$ is the element of order $r$ over $\operatorname{GF}(p^q)$. It can be proved that $\gamma_i$'s are distinct elements.

Another method to obtain solutions of $(1)$ is that considering the following equation:

$$ x^r=y^r \Rightarrow (x-y)\,g(x,y)=0 \tag{3} $$ then find roots of the $g(x,y)$ over $\operatorname{GF}(p^q)$, where $g(x,y)$ is a polynomial of degree $r-1$ and based on the variables $x$ and $y$.

My question: How to prove that $g(x,y)=\prod_{i=1}^{r-1} (x-\gamma^i\, y)$?

My try: We know by Newton's identities the following relation $$ \prod_{i=1}^{r-1} (x-\gamma^i\, y)=\sum_{k=0}^{r-1}(-1)^{r-1-k} e_{r-1-k}\,y^{r-1-k}\,x^k\tag{4} $$

In addition, it can be proved that for $1\leq k\leq r-1$, we have $$ p_k(\gamma^1,\cdots,\gamma^{r-1})=\sum_{i=1}^{r-1}\gamma^{ik}=-1\tag{5} $$ There is a relation between $p_k$ and $e_k$ by Newton's identities that results that $$ e_{2k}(\gamma^1,\cdots,\gamma^{r-1})=1 \quad, \quad e_{2k-1}(\gamma^1,\cdots,\gamma^{r-1})=-1 \quad, \quad 1\leq k \leq \frac{r-1}{2}\tag{6} $$ Therefore, by using $(5)$ and $(6)$ in the relation $(4)$ we conclude that

$$ \prod_{i=1}^{r-1} (x-\gamma^i\, y)=\sum_{k=0}^{r-1}y^{r-1-k}\,x^k=g(x,y) $$

Is it a correct proof?

Thanks for any suggestions.

Amin235
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    Looks ok to me. Good job learning Newton's identities - they show up surprisingly often. As does the sum $(5)$. – Jyrki Lahtonen Aug 19 '17 at 21:17
  • @JyrkiLahtonen I learned how to work by Newton's identities by some of your answers. The relation $(5)$, is proved by your comment in this question. I really thanks not just for answer of this question, but also for every note of finite field that I learnt from you. – Amin235 Aug 19 '17 at 21:22

1 Answers1

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I would just use the factorization $$f(x)=x^r-1=\prod_{i=0}^{r-1}(x-\gamma^i).$$ It implies that $$ x^r-y^r=y^rf(x/y)=y^r\prod_{i=0}^{r-1}(x/y-\gamma^i)=\prod_{i=0}^{r-1}(x-y\gamma^i). $$ You can then cancel the factor $x-y$ corresponding to $i=0$.

The trick is known as homogenization. It adds one more variable to a polynomial, and gives, as an end product, a homogeneous polynomial, i.e. a polynomial such that all the terms share the same total degree.

Jyrki Lahtonen
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