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Problem:
Let $a:\mathbb N \rightarrow \mathbb C $ a function with the property: $$\displaystyle{\sum_{d\mid n}a(d)=2^n, \forall n \in \mathbb N}.$$ Prove that $n\mid a(n), \forall n \in \mathbb N.$

My thought:

I used Möbius inversion formula to find that: $$\displaystyle{a(n)=\sum_{d\mid n}μ(n/d)2^d},\forall n \in \mathbb N.$$ From this formula we get information for the binary representation of $a(n)$. Then I thought to consider the binary representation of $n$, see if there is any connection between the binary representations of $a(n)$ and $n$ and if this connection can imply the desired relation of divisibility. I don't know if this idea can give me anything and how to proceed with it.

Any hints and ideas are highly appreciated.

Thanks in advance.

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    Now you should represent your $n$ as $p_{1}^{\alpha_{1}}\dots$ and consider all possible values of Mobius function. And considered $\frac{a(n)}{n}$ – 2017-01-13
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    @openspace It is a natural thought in number theory problems, but how is it gonna help here? – 2017-01-13
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    You wil get a sum of $2^{i}$ for some $i$ and this is a binary number with $1$ on $i$-th position, it's easier to consider this. – 2017-01-13
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    @openspace I know that. I can't see your point. – 2017-01-13

2 Answers 2

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I find if slightly more conofrable to use $a(n)=\sum\limits_{d|n}\mu(d)2^{n/d}$

Let $n$ be an integer and let $p^k$ be the maximum power of $p$ dividing $n$.

We have to prove $p^k$ divides $a(n)$.

To do this we use the inversion formula. Notice that $d$ can be viewed as subset of the prime divisors of $n$.

Notice that if $p\nmid d$ then $2^{n/d}\equiv 2^{n/dp}\bmod p^k$ because $n/d$ and $n/dp$ are both multiples of $p^{k-1}$ and they are congruent $\bmod p-1$ ( because $p$ is congruent to $1\bmod p-1$). It follows that $n/d$ and $n/dp$ are congruent $\bmod \varphi(p^k)$.

However, notice that $\mu(d)$ and $\mu(dp)$ have opposite signs, so we have shown that the whole sum cancels out $\bmod p^k$.

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    the case $p=2$ is still missing though. – 2017-01-13
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    but that case is easy, suppose $2^k$ is the largest power of $2$ dividing $n$. Notice $2^{k-1}$ is greater than or equal to $k$, and the smallest power of $2$ that you have to add is always at least $2^{2^{k-1}}\geq 2^n$ – 2017-01-13
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    I like the way you tackled it. Do you think that the approach with the binary representations can give an other solution? – 2017-01-13
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    I don't think that there is a "direct" approach that only uses the binary representation. I think you need to use something similar to euler's theorem somewhere down the road. – 2017-01-13
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    I didn't think of the p-adic valuation of $n$ and $a(n)$ and so I didn't try to tackle the problem with modulo $p^k$. Thus I couldn't handle the distinct powers of $2$. My bad. – 2017-01-13
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    Oh, I hadn't thought of it that way. – 2017-01-13
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Combinatorial proof.
Let $b(n)$ be the number of non-periodic binary sequences of length $n$. It is easily verified that $b(n)$ satisfies the same relation as $a(n)$: each of the $2^n$ binary sequences of length $n$ has a minimal period $d$ dividing $n$, and there are precisely $b(d)$ sequences with period $d$.

Thus from $a*1=b*1$ we have $$a(n)=b(n)$$ and because each non-periodic sequence of length $n$ comes with $n$ distinct cyclic permutations, $$n\mid b(n).$$

Remarks. 1. The same proof works with $2$ replaced by any positive integer; nowhere did we use any particular property of $2$.
2. This provides a proof of Fermat's little theorem, because $a(p)=2^p-2$ (with $2$ replaced by anything you want).

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    Which leaves to the question what happens with $2$ replaced by a negative integer... – 2017-01-14
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    Thanks for this proof. It is amazing! No Mobius inversion, but a combinatorial proof based on the binary representations, like I wanted. I would give you a second (+1) for the second remark, but I already gave you one. – 2017-01-14
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    @SachpazisStelios If you want to make me happy (and feel bad at the same time, because Jorge is my friend), you can accept this :) Do as you want... – 2017-01-14
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    In my opinion your solution is nicer, but I think it's not nice to uncheck Jorge's answer now. Next time I will wait more time before accepting a certain answer. I hope we are all OK. :) – 2017-01-14
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    As you wish. (That's why I usually wait a few days to accept an answer.) – 2017-01-14
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    You are right. I mentioned this in my edited comment above. – 2017-01-14
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    Let $a(n,k)$ be such that $\sum_{d\mid n}\,a(n,k)=k^n$ for all $n\in\mathbb{N}$ and $k\in\mathbb{Z}$. It can be shown that $a(n,k)$ is a polynomial of degree $n$ in $k$ with integer coefficients. Thus, if $n$ divides $a(n,k)$ for all positive integers $k$, then it follows that $n$ divides $a(n,k)$ for all integers $k$. – 2017-01-14
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    @Batominovski indeed, I just wrote the same here: http://math.stackexchange.com/questions/2097222 – 2017-01-14