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6 Tensor product does not seem to make any sense here. I suppose direct product was intended.

Hi there, I have been studying the following set (in order to investigate the average of products of Ramanujan sums with some weights).:

$$A=\lbrace (n,m) \in \mathbb{Z}/q\mathbb{Z} \otimes times \mathbb{Z}/q\mathbb{Z} | \mid \text{ }nm\equiv a \mod{q},\text{ } n \equiv 1 \mod{d_{1}}, \text{ } m \equiv 1 \mod{d_{2}} \rbrace$$

where q $q$ is any positive integer, $d_{1}$ and $d_{2}$ are proper divisors of $q$ and $(a,q)=1$.

for

For the case $a=1$, using a bit algebra I have deduced the elementary formula

$$\sum_{\substack{n_{1}n_{2}\equiv 1\mod{q}, \\ n_{1} \equiv 1 \mod{d_{1}}, \\ n_{2} \equiv 1 \mod{d_{2}}}}=\frac{\phi(q)}{lcm(\phi(d_{1}),\phi(d_{2}))}$$.

is mod{d_{2}}}}=\frac{\phi(q)}{\mathrm{lcm}(\phi(d_{1}),\phi(d_{2}))}.$$Is it possible to derive a formula for |A|? 5 improved formatting; added 14 characters in body Hi there, I have been studying the following sum set (in order to investigate the average of products of Ramanujan sums with some weights). \sum_{\substack{n_{1}n_{2}\equiv a\mod{q},  A=\lbrace (n,m) \in \n_{1mathbb{Z}/q\mathbb{Z} \otimes \mathbb{Z}/q\mathbb{Z} | \text{ }nm\equiv a \mod{q},\text{ } n \equiv 1 \mod{d_{1}}, \\ n_{2text{ } m \equiv 1 \mod{d_{2}}}} mod{d_{2}} \rbrace$$

where q is any positive integer, $d_{1}$ and $d_{2}$ are proper divisors of $q$ and $(a,q)=1$.

for the case $a=1$, using a bit algebra I have deduced the elementary formula

$\sum_{\substack{n_{1}n_{2}\equiv$\sum_{\substack{n_{1}n_{2}\equiv 1\mod{q}, \\ n_{1} \equiv 1 \mod{d_{1}}, \\ n_{2} \equiv 1 \mod{d_{2}}}}=\frac{\phi(q)}{lcm(\phi(d_{1}),\phi(d_{2}))}$. But evaluation of the above sum mod{d_{2}}}}=\frac{\phi(q)}{lcm(\phi(d_{1}),\phi(d_{2}))}$$. is it possible to derive a formula for the remaining cases are problematic.$|A|$? 4 added 4 characters in body Hi there, I have been studying the following sum (in order to investigate the average of products of Ramanujan sums with some weights).$\sum_{\substack{n_{1}n_{2}\equiv a\mod{q}, \\ n_{1} \equiv 1 \mod{d_{1}}, \\ n_{2} \equiv 1 \mod{d_{2}}}}$where q is any positive integer,$d_{1}$and$d_{2}$are proper divisors of$q$and$(a,q)=1$. for the case$a=1$, using a bit algebra I have deduced the elementary formula$\sum_{\substack{n_{1}n_{2}\equiv 1\mod{q}, \\ n_{1} \equiv 1 \mod{d_{1}}, \\ n_{2} \equiv 1 \mod{d_{2}}}}=\frac{\phi(q)}{lcm(\phi(d_{1}),\phi(d_{2}))}\$.

But evaluation of the above sum for the remaining cases are problematic.

3 edited body; added 3 characters in body; edited body
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