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

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edited Feb 12 at 14:15

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 $|A|$?

improved formatting; added 14 characters in body

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edited Feb 11 at 19:40

Egeselazuzyan
11●2

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.$|A|$?

added 4 characters in body

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edited Feb 11 at 18:10

Egeselazuzyan
11●2

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.