Question

Dados m e a naturais, mostre que se m | a - 1 então,

$\mathsf{mdc\Big(\dfrac{a^m-1}{a-1}, a - 1\Big)=m}$


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Por favor reponder de forma detalhada.

Answer 1

Como $m~|~a-1$, então $a-1=mq$, sendo $q\in\mathbb{N}$.

Assim, $a=mq+1$ e $a^{m}-1=(mq+1)^{m}-1$ e $\dfrac{a^{m}-1}{a-1}=\dfrac{(mq+1)^{m}-1}{mq}$.

Veja que $(xy+1)^3=x^3y^3+3x^2y^2+3xy+1=xy(x^2y^2+3xy+3)+1$.

Disso, deduzimos que:

$(mq+1)^{m}-1=mq\cdot\left[(mq)^{m-1}+\dbinom{m}{1}(mq)^{m-2}1^1+\dbinom{m}{2}(mq)^{m-3}1^2+\dots+m\right]$

Logo:

$\dfrac{a^{m}-1}{a-1}=\dfrac{mq\cdot\left[(mq)^{m-1}+\dbinom{m}{1}(mq)^{m-2}1^1+\dbinom{m}{2}(mq)^{m-3}1^2+\dots+m\right]}{mq}$

$\dfrac{a^{m}-1}{a-1}=(mq)^{m-1}+\dbinom{m}{1}(mq)^{m-2}1^1+\dbinom{m}{2}(mq)^{m-3}1^2+\dots+m$

$\dfrac{a^{m}-1}{a-1}=m\cdot\left[m^{m-2}q^{m-1}+\dbinom{m}{1}m^{m-3}q^{m-2}1^1+\dbinom{m}{2}m^{m-4}q^{m-3}1^2+\dots+1\right]$

Lembrando que $a-1=mq$ concluímos que:

$\boxed{\text{mdc}\left(\dfrac{a^{m}-1}{a-1},a-1\right)=m}$