**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**.

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...

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

integralperiod 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...

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.

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...