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If ${q^k}\sigma(q^k) \equiv a \pmod b$ and $\displaystyle\frac{\sigma(q^k)}{n} \neq \displaystyle\frac{\sigma(n)}{q^k}$, does it follow that $n\sigma(n) \not\equiv a \pmod b$?

Here, $q$ is prime and $n$ is composite.

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up vote 2 down vote accepted

We can generate counterexamples at will. Take for example $n=4$. Then $n\sigma(n)=28$.

Take $q=11$, $k=1$. Then $q^k\sigma(q^k)=132$. We find $a$ and $b$ such that $28\equiv a \pmod{b}$ and $132\equiv a\pmod{b}$. Choose $a=2$. Then $28\equiv 2\pmod{13}$ and $132\equiv 2\pmod{13}$. It is too easy to arrange for two numbers to be each congruent to $a$ modulo $b$ if we are allowed to choose $a$ and $b$. A simpler class of counterexamples would use $b=1$.

Added: For an example with $a=2$, $b=4$ as asked for in a comment, let $q=17$, $k=1$, and $n=15$. Then $q^k\sigma(q^k)\equiv 2\pmod{4}$ but $n\sigma(n)\not\equiv 2\pmod{4}$.

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@ArnieB.Dris: Sure, $n=15$, then $\sigma(n)=24$ so $n\sigma(n)\equiv 0\pmod{4}$. Now arrange in any way for $q^k\sigma(q^k)$ to be congruent to $2$ modulo $4$. Like $q=17$, $k=1$. There is too much freedom in the problem. – André Nicolas Jun 22 '12 at 14:33
Yes, I had a typo in the comment, $n\sigma(n)\not\equiv a \pmod{b}$, where $a=2$ and $b=4$, as requested. The middle condition about the ratio imposes no congruential restriction, might as well not be there. – André Nicolas Jun 22 '12 at 14:38
@ArnieB.Dris: There is undoubtedly a real problem underneath, it is just a question of finding the right formulation. – André Nicolas Jun 22 '12 at 14:43
@ArnieB.Dris: I have problems that I return to from quite a number of years before that. Someday $\dots$! – André Nicolas Jun 22 '12 at 14:56
@ArnieB.Dris: Comments are not the right medium, not enough space to ask a precise question. As you know, for $n$ odd composite $\sigma(n)\equiv 2\pmod{4}$ is very restrictive. We need $n=p^t$, $p\equiv 1\pmod{4}$, $t\equiv 1\pmod{4}$. So let $n=5^5$, or $17^9$, or $\dots$. – André Nicolas Jun 22 '12 at 15:07

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