Let $\sigma$ denote the classical sum-of-divisors function.
What is $\gcd(\sigma(q^{k-1}), \sigma(q^k))$?
Update: I have transferred the transcript of my attempt to an actual answer to this MSE question. Thanks to mixedmath for pointing it out!
2026-03-29 14:58:53.1774796333
What is $\gcd(\sigma(q^{k-1}), \sigma(q^k))$?
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Here is my attempt at an answer:
$$\sigma(q^{k-1}) = \frac{q^k - 1}{q - 1}$$ $$\sigma(q^k) = \frac{q^{k+1} - 1}{q - 1}$$
Therefore:
$$\gcd(\sigma(q^{k-1}),\sigma(q^k)) = {\left(\frac{1}{q - 1}\right)}\gcd(q^k - 1, q^{k+1} - 1) = \frac{q^{\gcd(k,k+1)} - 1}{q - 1} = 1.$$