# A sum of Ramanujan sums

I have the following question about Ramanujan sums.

(All vectors and matrices here will be understood to have integer entries.)

Let $X_q=\{ (x_1,...,x_R)|1\leq x_i\leq q\}$ and let, for any $R\times R$ matrix $B$ with rows $\mathbf b_i$, $c_q(B\mathbf x):=c_q\left (\mathbf b_1\cdot \mathbf x\right )...c_q\left (\mathbf b_r\cdot \mathbf x\right ),$ where $c_q(n)$ is Ramanujan's sum. Suppose an $R\times R$ matrix $A$ with determinant $D$ is given and denote by $I$ the identity matrix. Then my question would be:

Does

$S=\sum _{\mathbf x\in X_q\atop {A\mathbf x\equiv \mathbf 0(q)}}c_q(I\mathbf x)$

depend only on $(q,D)$?

In fact I thought the sum would be zero except for the case (q,D)=1 but I'm not so sure this is true.

Why might it be true? Write $d_i$ for the $i$-th elementary divisor of $A$. Then on considering the Smith normal form of $A$ and after using $c_{dq}(dn)=f(q,d)c_q(n)$ we have, up to a function depending only on $q$ and $(q,d_i)$,

$S=\sum_{x\in X_{(q,D)}}c_{(q,D)}\left (QE\mathbf x\right ),$

where $Q=Q(A)$ is invertible and $E=(e_{i,j})$ is a diagonal matrix with

$e_{i,i}=\frac {(q,d_R)}{(q,d_i)}.$

It seems to me plausible that this last sum is zero except for when $(q,d_R)=1$, which would be perfect. Testing thousands of $3\times 3$ matrices gives me always zero, so I do start to believe it. Could it be true? And along which lines should I be thinking if I want to prove it?

Any suggestions would be very much appreciated!

• Ah, it should be interpreted as the identity matrix times vector x... it's unclear though, I'll edit Nov 3, 2016 at 14:11