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For a non-negative integer $\ell$, define the arithmetic functions $\sigma_{\ell}(k) = \sum_{d \mid k} d^{\ell}$ and $\bar{\sigma}_{\ell}(k) = \sum_{k = dd^{\prime}} (-1)^{(d^{\prime} - 1)/2} d^{\ell}$.

The function $\sigma(k) = \sigma_{1}(k)$ is the divisor-sum function.

Question: Let $k$ be an odd integer. Is the following congruence known?

\begin{align*} 3\sigma(k) &\equiv \sigma_{3}(k) + 2 \bar{\sigma}_{0}(k) &\pmod{16} \end{align*}

Thanks!

Update: What about this one? \begin{align*} \bar{\sigma}_{2}(k) &\equiv \bar{\sigma}_{0}(k) &\pmod{4} \end{align*}

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This is not actually a statement about divisor sums; it's just a sum of the corresponding statements for the individual divisors:

$3d \equiv d^3 + 2 (-1)^{(d-1)/2}\pmod{16}\; \mbox{for odd } d\;.$

If you substitute $4n+d$ for $d$ in this equation, $n$ drops out, so you only have to check the two cases $d=1$ and $d=3$.

Update in response to the update in the question:

Again, that result is just simpler results in a complicated "disguise". In fact $\bar{\sigma}_{2}(k) \equiv \bar{\sigma}_{0}(k) \pmod{4}$, since $d^2 \equiv d^0\pmod{4}$ for the individual odd divisors. That you can stick an additional sign factor in is due to the fact that it doesn't make a difference if the number of divisors is even, only squares have an odd number of divisors and the sign factor is $1$ for $k$ an odd square.

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    @Derek Jennings: Agreed.2011-02-12