Where m is the largest odd factor of n, I am trying to prove that $$\sigma(n)\equiv d(m)\mod{2}$$
Any help is appreciated!
2026-03-26 12:51:36.1774529496
Mods and Multiplicative functions
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I presume $\sigma(n)$ is the sum of the divisors of $n$. Then $\sigma(n)\equiv\sigma(m)\pmod2$ as one can ignore even divisors of $n$. For odd $m$, $\sigma(m)\equiv d(m)\pmod2$ as all divisors of $m$ are odd.