# Bezout's identity in finite fields

Suppose $x_1 , x_2 \in \Bbb{F}_{p^n}^*$. Let $f_i$ be the order of $x_i$ in the cyclic group $\Bbb{F}_{p^n}^*$. Using Bezout's identity we find an $\alpha$ and a $\beta$ such that $\alpha f_1 + \beta f_2=gcd(f_1,f_2)$. Suppose $x=x_1^\beta x_2^\alpha$. How do we prove that $x^{\frac{lcm(f_1,f_2)}{f_i}} = x_i$ for $1\leq i \leq 2$?

I tried by finding a generator $\omega$ of the cyclic group. Then $x_1=\omega^{k_1}$ for a certain $k_1$ and $x_2=\omega^{k_2}$ for a certain $k_2$. Thus we have $x=\omega^{(\alpha k_1 + \beta k_2)}$. We also have that $lcm(f_1,f_2)=\frac{f_1*f_2}{gcd(f1,f2)} = \frac{f_1*f_2}{\alpha f_1 + \beta f_2}$. So $x^{\frac{lcm(f_1,f_2)}{f_1}}= x^{\frac{f_2}{\alpha f_1 + \beta f_2}} = \omega^{\left(\frac{f_2(\alpha k_1+\beta k_2)}{\alpha f_1 + \beta f_2}\right)}$. How do I continue ? or is this the wrong direction ?

• As stated, the claim is false I think? If $x_1, x_2$ have equal order but different values, then the exponent is the same for both. Consider the theorem with $x_1= a, x_2=a+1$ with $p=2, n=2$. Both have order 3, so the LCM is 3, and the exponent is the same for all i, so we get $a=a+1$, which is a contradiction. – Artimis Fowl May 7 '17 at 18:48

The following is a partial solution - I believe the claim is missing some constraint on $f_i$ or possibly $i$.

One can get close with just the given $x = x_1^\alpha x_2^\beta$. Note that $d = gcd(f_1, f_2)$ divides $lcm(f_1, f_2), f_1, f_2$. Let's assume WLOG $i=1$. Note:

$$\frac{lcm(f_1,f_2)}{f_1} = \frac{f_2}{d}$$

So we can simplify the exponent, and use our defined value of $x$:

$$x_1^{\beta f_2 / d} x_2^{\alpha f_2 / d} = x_1^{1 - \alpha f_1 / d} x_2^{\alpha f_2/d} = x_1 \frac{x_2^{\alpha f_2 / d}}{x_1^{\alpha f_1 / d}}$$

The second step there used $d = \alpha f_1 + \beta f_2 \implies d - \alpha f_1 = \beta f_2$, and distributed the $/d$. The last equality is just rearranging.

Now, we're finished if we can show $\frac{x_1^{\alpha f_1 / d}}{x_2^{\alpha f_2 / d}} = \sqrt[d]{1} = 1$. Ie, that $y^d = 1$ has only one solution in $\mathbb{F}_{p^i}$. Unfortunately this needn't be the case, take for example $\alpha, 1+ \alpha$ in $\mathbb{F}_4$. $d=3$, and cube roots are not unique. In fact the ending equation need not satisfy...