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I know that $\frac{a^n - 1}{a-1}$ is a geometric series, but I don't know how that can help me solve it.

This is not a duplicate question...

My Name
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  • @labbhattacharjee This is not a duplicate because the special case has a simpler proof than the question you linked to. – David Feb 03 '16 at 09:15

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We have $a \equiv 1 \pmod{a - 1}$. Therefore $$\frac{a^n - 1}{a - 1} = 1 + a + \dots + a^{n-1} \equiv 1 + 1 + \dots + 1 = n \pmod{a - 1}.$$ Hence $\gcd[(a^n-1)/(a-1),a-1] = \gcd(n,a-1)$.

David
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Notice that $$\gcd(\frac{a^n-1}{a-1},a-1)=\gcd(a^{n-1}+\cdots+a+1,a-1)$$ Now we have $k=1\cdot (a-1)+2a\cdot (a-1)+3a^2\cdot (a-1)+\cdots +(n-1)a^{n-1}(a-1)$, so that $a-1|k$, and so $$\gcd(a^{n-1}+\cdots+a+1,a-1)=\gcd(a^{n-1}+\cdots+a+1-k,a-1)=\gcd(na^{n-1},a-1)$$ Now since $\gcd(a^{n-1},a-1)=1$, we know $$\gcd(\frac{a^n-1}{a-1},a-1)=\gcd(n,a-1)$$