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    #11 Thread Author
    Quote Originally Posted by I like Serena View Post
    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?


    Quote Originally Posted by I like Serena View Post
    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...

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    #12 Thread Author
    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 $0$ ?

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

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    #14
    Quote Originally Posted by evinda View Post
    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?

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    #15
    Quote Originally Posted by evinda View Post
    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.

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