# Thread: Result of Euclidean division

Originally Posted by I like Serena
Well... if we divide $x^6-1$ by $x-\alpha^{a+1}$ first, we get:
$$\frac{x^6-1}{x-\alpha^{a+1}} = x^5 + \alpha^{a+1}x^4 + \alpha^{2(a+1)}x^3 + \alpha^{3(a+1)}x^2 + \alpha^{4(a+1)}x + \alpha^{5(a+1)} +\frac{\alpha^{6(a+1)} - 1}{x-\alpha^{a+1}}$$
Note that the remainder term $\alpha^{6(a+1)} - 1=0$, because $\alpha$ is a 6-th root of unity.

That leaves dividing what we found by $x-\alpha^{a+2}$...
So we divide $x^5 + \alpha^{a+1}x^4 + \alpha^{2(a+1)}x^3 + \alpha^{3(a+1)}x^2 + \alpha^{4(a+1)}x + \alpha^{5(a+1)}$ by $x- \alpha^{a+2}$ , right?

I got that $$x^5 + \alpha^{a+1}x^4 + \alpha^{2(a+1)}x^3 + \alpha^{3(a+1)}x^2 + \alpha^{4(a+1)}x + \alpha^{5(a+1)}=(x- \alpha^{a+2}) (x^4+ \alpha^{a+1}(\alpha+1) x^3+ \alpha^{2a+2}[\alpha^2+ \alpha-1] x^2+ \alpha^{3a+3}[1+ \alpha [\alpha^2+ \alpha-1]]x+ \alpha^{4a+4}[1+ \alpha(1+ \alpha(\alpha^2+\alpha-1) )])+ \alpha^{5a+5}[\alpha [1+ \alpha (1+ \alpha(\alpha^2+ \alpha+1))]]$$

Is it again wrong?

Originally Posted by I like Serena
Alternatively, if I redo your calculation, I get the same terms for $x^4$, $x^3$, and $x^2$, but a different term for $x$.
Mine is $\alpha^{a+1}(\alpha^{2a+2} - 2\alpha^{a+2}) x = \alpha^{2a+3}(\alpha^a - 2)x$.
Oh, and my final remainder term has a constant of $1+\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5$, which is indeed $0$. It suggests that my calculation is correct.

Perhaps there is a mistake in your calculation.
Ok, I will retry them...

Again I get the same result as before...

Also why does it hold that $1+\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5$ is equal to $0$ ?

So what did you find $\frac{x^6-1}{x^2- \alpha^{a+1} (\alpha+1) x+ \alpha^{2a+3}}$ to be equal to?

4. Originally Posted by evinda
Also why does it hold that $1+\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5$ is equal to $0$ ?
Isn't it a geometric series? What would its sum be?

5. Originally Posted by evinda
So what did you find $\frac{x^6-1}{x^2- \alpha^{a+1} (\alpha+1) x+ \alpha^{2a+3}}$ to be equal to?
Let $B=\alpha^{a+1}(\alpha+1)$ and $C=\alpha^{2a+3}$.

Then:
\begin{aligned}x^6-1 &= \Big(x^2-Bx+C\Big)\
\Big(x^4 + Bx^3 + (B^2-C)x^2 + B(B^2-2C)x + (B^4 - 3B^2C + C^2)\Big) \\
&+ \Big[B(B^4-3B^2C+C^2) - BC(B^2-2C)\Big]x - C(B^4-3B^2C+C^2) - 1
\end{aligned}

I veried that $- C(B^4-3B^2C+C^2) - 1 = 0$ by filling in $\alpha$ and making use of its properties.
I also filled in $\alpha$ in $Bx^3$ and in $(B^2-C)x^2$ and found the same thing you did.