Prove, $\gcd((a^b-1)/(a-1), (a^c-1)/(a-1))=1$

Question

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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Answer 1

Let $f(n) = (a^n-1)/(a-1).\,$ By this answer $\,(f(b),f(c)) = f((b,c)) = f(1) =1$

Answer 2

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 $y$.