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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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I don't understand, why is there a $3$ in the $mod$ $2$ congruence? –  Eric Naslund Feb 12 '11 at 5:55
    
Corrected. Thanks. –  user02138 Feb 12 '11 at 15:36

1 Answer 1

up vote 4 down vote accepted

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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Although the congruences for odd $d$ are true, the question contains the function $\bar{\sigma}_0(n)$ and so we do actually need to sum up over the divisors and note that $\bar{\sigma}_0(n)= \sum_{d|n}(-1)^{(d^{\prime}-1)/2}= \sum_{d|n}(-1)^{(d-1)/2}$ to finish off the argument, so perhaps your initial statement is a little misleading. The OP asks whether it's known, rather than if it's true (though admittedly one could read this as, "known to be true"). Given your observation, it's probably not significant enough to be of any special interest and thus acquire the status of a "known result". –  Derek Jennings Feb 12 '11 at 12:23
    
@Derek Jennings: I don't understand the thrust of your comment. $\bar{\sigma}_0$ doesn't contain $d$ at all, so it's also just a simple sum over the divisors; all there is to note is that it doesn't matter whether we call the divisor $d$ or $d'$ -- so whatever is more here than a statement about individual divisors is just "notational disguise"; I don't see in what sense we need to "sum up over the divisors". I agree with the second part of your comment, though. –  joriki Feb 12 '11 at 12:37
    
@Derek Jennings: I see now that you wrote $\bar{\sigma}_0$ as a sum over $d\mid n$ -- that's not how it's defined in the question, where $d$ and $d'$ play symmetric roles and it's more obvious than it looks in your comment that $\bar{\sigma}_0$ is just a sum over $d'$. –  joriki Feb 12 '11 at 12:41
    
@joriki. There actually is a prime in the first exponent in my comment, it's just very small! And that's my point, $\bar{\sigma}_0(n)$ contains $d^{\prime}$ and we do NOT have $3d \equiv d^3 + 2 (-1)^{(d^{\prime}-1)/2}\pmod{16}.$ ($n=35,d=5,d^{\prime}=7,$ for example) That's why I think it's worth mentioning $\bar{\sigma}_0(n)= \sum_{d|n}(-1)^{(d^{\prime}-1)/2}= \sum_{d|n}(-1)^{(d-1)/2} .$ –  Derek Jennings Feb 12 '11 at 13:15
    
@Derek Jennings: I think that was a misunderstanding -- I did see the prime in your comment; I was commenting on the fact that you wrote $\bar{\sigma}_0$ as a sum over $d \mid n$ whereas in the question it's written as a sum over $dd'=n$. The way you write it, you do need the sum and an argument that each $d':=n/d$ occurs exactly once in the sum, whereas the way it's written in the question all you need to do is rename $d'$ to $d$ and drop all the summation signs. –  joriki Feb 12 '11 at 13:29

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