For positive integers $s$ and $n$ , the complex number $$c_s(n)=\sum_{\substack{0\le k<n \\ (k,n)=1}}e^{2\pi iks/n}$$ is referred to as a Ramanujan sum, or a Ramanujan trigonometric sum. Since $e^{2\pi i}=1$ , an equivalent definition is $$c_s(n)=\sum_{k\in r(n)}e^{2\pi iks/n}$$ where $r(n)$ is some reduced residue system mod $n$ , meaning any subset of $\mathbb{Z}$ containing exactly one element of each invertible residue class mod $n$ .

Using a symmetry argument about roots of unity, one can show

Using the Chinese remainder theorem, it is not hard to show that for any fixed $n$ , the function $s\mapsto c_s(n)$ is multiplicative: $$c_s(n)c_t(n)=c_{st}(n) \text{ if }(s,t)=1 \;.$$

If $m$ is invertible mod $n$ , then the mapping $k\mapsto km$ is a permutation of the invertible residue classes mod $n$ . Therefore $$c_s(mn)=c_s(n)\text{ if }(m,s)=1\;.$$ Remarks: Trigonometric sums often make convenient apparatus in number theory, since any function on a quotient ring of $\mathbb{Z}$ defines a periodic function on $\mathbb{Z}$ itself, and conversely. For another example, see Landsberg-Schaar relation.

Some writers use different notation from ours, reversing the roles of $s$ and $n$ in the expression $c_s(n)$ .

The name ``Ramanujan sum'' was introduced by Hardy.