**Update**:


I have found reference to this problem. It is known as "the Rédei-de Bruijn-Schönberg theorem", which is proved in the following papers:


* N. G. de Bruijn: *On the factorization of cyclic groups*, Indag. Math.15(1953), 370-377.
* L. Rédei: *Ein Beitrag zum Problem der Faktorisation von Abelschen Gruppen*, ActaMath. Acad. Sci. Hungar.1(1950), 197-207.
* I. J. Schoenberg: *A note on the cyclotomic polynomial*, Mathematika11(1964), 131-136.


**End of update**.




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The background of this question is something about cyclotomic fields, but the statement doesn't involve any algebraic number theory. I just get puzzled by this (might be stupid) little question...




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Let $n>1$ be an integer, and consider the vector space $\mathbb{C}^n$.


A vector $v=(v\_1, \cdots, v\_n) \in \mathbb{C}^n$ is called


* *periodic*, if there is a proper divisor $d$ of $n$, such that $v\_i = v\_{i + d}$ for all $i$;
* *integral*, if every $v\_i$ is an integer.



> 
> Question: if an integral vector can be written as a finite sum of periodic vectors, then is it true that it can always be written as a finite sum of **integral** period vectors?
> 
> 
> 


It should be clear that the field $\mathbb{C}$ could be replaced with any field of characteristic zero (e.g. $\mathbb{Q}$).


I would guess that the claim is true, but cannot convince myself with a proof...




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So far I can only prove the case when $n$ has at most $2$ different prime factors, which doesn't help much in the general case.


I've also tried to adopt a point of view from cyclotomic fields, or representation theory, or doing some Fourier transform - but again I'm not intelligent enough to morph the question to something known...


Therefore I add all the possibly relevant tags.




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EDIT: I add here a proof when $n = pq$ is the product of $2$ different primes. The more general case $n = p^r q^s$ is morally the same.


Since $n = pq$, every periodic vector $a$ either satisfies $a\_i = a\_{i + p}$ for all $i$, or satisfies $a\_i = a\_{i + q}$ for all $i$.


Hence any finite sum of periodic vectors can be written as $a + b$, where $a\_i = a\_{i + p}$ and $b\_i = b\_{i + q}$.


Now suppose that $v = a + b$ is such a sum, which is integral. This means that $a\_i + b\_i = 0\mod\mathbb{Z}$, which then gives:
$$a\_i = -b\_i = -b\_{i + q} = a\_{i + q} \mod \mathbb{Z}.$$
Together with $a\_i = a\_{i + p}$, we conclude that all $a\_i$ are equal $\mod\mathbb{Z}$, hence all $b\_i$ also, and we can adjust them with a constant vector so as to make them both integral.


This trick doesn't work for more than $2$ prime factors, though...



Is an integral sum of periodic vectors always a sum of integral periodic vectors?
A beautiful question! Though I don't have the time or the space to fill in all the details, I think one can answer the question in the following way. The answer is yes.


We can reformulate the question as follows: let $I \subset \mathbb{Z}[x]$ be the ideal generated by the polynomials
$$
\frac{x^n-1}{x^d-1}, \quad d \mid n, \quad d \neq n.
$$
We claim that $I$ is *$\mathbb{Z}$-saturated,* that is, within $\mathbb{Q}[x]$, 
$$
\mathbb{Q} I \cap \mathbb{Z}[x] = I.
$$
(Attach to each polynomial $f \in \mathbb{Q}[x]$ the sequence given by the coefficients of $f$ mod $x^n - 1$. The passage from $\mathbb{C}$ to $\mathbb{Q}$ is effected by some easy linear algebra.)


We claim, indeed, that 
$$
I = J := (\Phi\_n)
$$
is the ideal generated by the $n$th cyclotomic polynomial. Note that $J$ is visibly $\mathbb{Z}$-saturated.


1. Clearly
$$
(x^n-1) \subseteq I \subseteq J.
$$
2. Also, $\mathbb{Q}I = \mathbb{Q}J$, because the only common roots of the generators of $I$ are at the primitive $n$th roots of unity and are all simple. (This step uses the Nullstellensatz, which, since we have just one variable, isn't all that deep.)
3. So if $I \neq J$, the discrepancy can be found locally, that is, there is a prime $p$ such that the images $\bar{I}, \bar{J}$ of $I$ and $J$ in $\mathbb{F}\_p[x]$ are unequal. (Pass to the finite-dimensional $\mathbb{Q}$-vector space $\mathbb{Q}[x]/(x^n-1)$. The images $L\_I$, $L\_J$ are $\mathbb{Z}$-lattices with the same $\mathbb{Q}$-span; if they are unequal, pick $p | [L\_J : L\_I]$.)
4. If $p \nmid n$, we are done by the Nullstellensatz again, because the $n$th roots of unity remain distinct in $\bar{\mathbb{F}}\_p$.
5. Now $p \mid n$. Let
$$
\bar K = \left\{\frac{f}{\Phi\_n(x)} : f \in \bar I \right\}.
$$
We wish to prove that $\bar K$ is the unit ideal, that is, that the polynomials in $\bar K$ have no common roots. Suppose there is a common root $\lambda \in \bar{\mathbb{F}}\_p$. Then $\lambda^n = 1$, since $x^n - 1 \in \bar I$. Let $\lambda$ be a primitive $r$th root of unity; note that $r \mid n$ and $p \nmid r$, so $r \mid n'$, where $n = p^k n'$ with $p \nmid n'$. We get a contradiction as follows:


	* If $r \neq n'$, then the element
	$$
	\frac{x^n - 1}{\Phi\_n(x) (x^{p^k r} - 1)} = \prod\_{d|n,\, d\nmid p^k r,\, d \neq n} \Phi\_d \in \bar K
	$$
	has no root at $\lambda$ (because the factors $\Phi\_r, \Phi\_{pr}, \ldots, \Phi\_{p^k r}$ have all been eliminated).
	* If $r = n'$, then the element
	$$
	\frac{x^n - 1}{\Phi\_n(x) (x^{n/p} - 1)} = \prod\_{d|n,\, p^k \parallel d,\, d \neq n} \Phi\_d \in \bar K
	$$
	likewise has no root at $\lambda$.


