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What is $\gcd(a^2-1, a^3-1)$? Is it $1$? The exponents seem to follow the pattern of $\gcd(a, a+1)$.

KReiser
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    Observe that if $a$ is odd then $2$ divides both, so answer cannot be $1$ in all cases. Moreover $a-1$ divides both, so.. – Anurag A Oct 18 '20 at 05:17
  • Hint: use the Euclidean algorithm: $!\bmod a^2-1!:\ {a^2\equiv 1}\Rightarrow a^3-1\equiv a-1.,$ Next $\bmod a-1!:\ a\equiv 1\Rightarrow, a^2-1\equiv 0,,$ so the gcd = $(0,a-1) = a-1.,$ Generally $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ – Bill Dubuque Oct 18 '20 at 05:27
  • More conceptually $,d\mid a^2!-!1,a^3!-!1\iff !\bmod d!:\ a^2\equiv 1\equiv a^3!\iff! a\equiv 1!\iff! d\mid a!-!1,,$ so $,{a^2!-!1,a^3!-!1},$ and $,{\color{#c00}{a!-!1}}$ have same set of common divisors $,d,$ so the same greatest common divisor, which clearly must be $,\color{#c00}{a!-!1}.\ $ This proof generalizes widely - see the linked post. – Bill Dubuque Oct 18 '20 at 05:47

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$a^3-1 = (a-1)(a^2+a+1) = (a-1)(a(a+1)+1)$

$a^2-1=(a-1)(a+1)$

${\rm GCD}(a+1,a(a+1)+1)={\rm GCD}(a+1,1)=1$ so that ${\rm GCD} (a^2-1,a^3-1)=a-1$

HK Lee
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