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Source. Theorem 1.3, PDF pp. 3--4 of arXiv:math/0409279v2, the copy read for this card.

Statement

Let $\{a_s(n_s)\}_{s=1}^k$ be the finite system (1.1) of the paper, with $k>1$ and positive integer moduli $n_s$, and let $\lambda_1,\ldots,\lambda_k\in\mathbb Z$ be weights attached to its $k$ residue classes. Put

$$w(x)=\sum_{\substack{1\leq s\leq k\\ n_s\mid x-a_s}}\lambda_s .$$

Let $m\in\mathbb Z$, and suppose that $n_0\in\mathbb Z^+$ is the smallest positive period of $w(x)$ modulo $m$. Let $d\in\mathbb Z^+$ be such that $d\nmid n_0$ and

$$I(d)=\{1\leq s\leq k: d\mid n_s\}\neq\emptyset .$$

Then either $m$ divides

$$[n_1,\ldots,n_k]\sum_{s\in I(d)}\frac{\lambda_s}{n_s},$$

where $[n_1,\ldots,n_k]$ is the least common multiple of the moduli, or

$$|I(d)|\geq\left|\{a_s\bmod d: s\in I(d)\}\right| \geq\min_{\substack{0\leq s\leq k\\ s\notin I(d)}}\frac{d}{(d,n_s)} \geq p(d),$$

where $(d,n_s)$ is the greatest common divisor and $p(d)$ is the smallest prime divisor of $d$.

The minimum runs over $0\leq s\leq k$, so it includes the index $s=0$ with modulus $n_0$, the least period; that index is never in $I(d)$, whose indices run from $1$ to $k$. The integer $m$ is not required to be positive. The source calls the theorem a refinement of Theorem 1.1, and its Remark 1.4 (PDF p. 4) attributes the case $m=0$ to the author's 1991 paper, with an extension in his 2004 paper.

Proof pointer. The proof is in Section 2, PDF pp. 5--6. It evaluates the weighted sum over one least period against roots of unity of order $d$, as in the proof of Theorem 1.1, and then uses a linear recurrence with algebraic-integer coefficients. The proof was not reconstructed or independently checked here.

Bears on. No Erdős problem directly.