Prove, $\gcd((a^b-1)/(a-1), (a^c-1)/(a-1))=1$ Problem  : Prove (or disprove) that,  $\gcd((a^b-1)/(a-1),  (a^c-1)/(a-1))=1$ (greatest common divisor), 
when $a, b, c $ are prime numbers and $a, b, c \geq 3, b \neq c$.  $(a^b-1),  (a^c-1)$ are factors of $(a^{bc}-1)$.
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assume $c > b$
Also gcd $(a, 1 + a + ... + a^b) = 1$
Let $x = (a^c - 1)/(a-1) = 1 + a + a^2 + ... + a^{c-1}$
Let $y = (a^b - 1)/(a-1) = 1 + a + a^2 + ... + a^{b-1}$
$gcd(x,y) = gcd(x-y, y) = gcd((a^b + ... + a^{c-1}), y) = gcd ((a^b(1 + a + ... + a^{c-b-1}), y) = gcd((1 + a + ... + a^{c-b-1}), y)$
as we keep doing, we get $gcd(x,y) = gcd((1 + a + ... + a^{k-1}), (1 + a + ... + a^{b-1}))$ with $k < b$ 
eventually ending up with $gcd(1 + a + ... + a^m, a^n) = 1$ for some integer $m$ and $n$. 
Therefore there is no common factor between $x$ and $y$.
