$$f(n)= \genfrac(){}{0}{-15}{n}$$ if $n \neq 1$ is odd.
What is $\sum_{d \mid n}f(d)$ ?
Here is what I tried : Let $n=p_1^{a_1} \cdot \cdot \cdot p_r^{a_r}$, for $p_1,...,p_r$ odd primes. If $p_1,...,p_r$ contains 3,5 then since $gcd(-15,5)=5,gcd(-15,3)=3$, the Jacobi symbols of $\genfrac(){}{0}{-15}{5}=\genfrac(){}{0}{-15}{3}=0$, hence we can deduce that $n$ cannot contain a prime factor of 3 or 5.
We can simplify $\genfrac(){}{0}{-15}{n}=\genfrac(){}{0}{-15}{p_1^{a_1}} \cdot \cdot \cdot \genfrac(){}{0}{-15}{p_r^{a_r}}=\genfrac(){}{0}{-1}{p_1^{a_1}} \cdot \cdot \cdot \genfrac(){}{0}{-1}{p_r^{a_r}}\genfrac(){}{0}{15}{p_1^{a_1}} \cdot \cdot \cdot \genfrac(){}{0}{15}{p_r^{a_r}}$
Then by a theorem for Legendre's symbol, $\genfrac(){}{0}{-1}{p_i^{a_i}}=(\genfrac(){}{0}{-1}{p_i})^{a_i}=(-1)^{a_i(p-1)/2}, \forall i$, since $(p-1)/2$ is odd, $((-1)^{(p-1)/2})^{a_i}=(-1)^{a_i}$
Thus the above becomes $(-1)^{a_1+\cdot\cdot\cdot+a_r}\genfrac(){}{0}{15}{p_1^{a_1}} \cdot \cdot \cdot \genfrac(){}{0}{15}{p_r^{a_r}}=(-1)^{a_1+\cdot\cdot\cdot+a_r}(\genfrac(){}{0}{15}{p_1})^{a_1} \cdot \cdot \cdot (\genfrac(){}{0}{15}{p_r})^{a_r}$
Which could be further simplified to $(-1)^{a_1+\cdot\cdot\cdot+a_r}(\genfrac(){}{0}{3}{p_1})^{a_1} \cdot \cdot \cdot (\genfrac(){}{0}{3}{p_r})^{a_r}(\genfrac(){}{0}{5}{p_1})^{a_1} \cdot \cdot \cdot (\genfrac(){}{0}{5}{p_r})^{a_r}$
Now as $3^{2}=9, 5^{2}=25$, both is not a square mod of an odd prime, $$(\genfrac(){}{0}{3}{p_r})^{a_r}(\genfrac(){}{0}{5}{p_1})^{a_1} \cdot \cdot \cdot (\genfrac(){}{0}{5}{p_r})^{a_r}=(-1)^{a_1}\cdot\cdot\cdot(-1)^{a_r}(-1)^{a_1}\cdot\cdot\cdot(-1)^{a_r}=1$$ So the expression above simplifies to just $(-1)^{a_1+\cdot\cdot\cdot+a_r}$.
Now if the above were to be correct, I am stuck at finding $\sum_{d \mid n}f(d)$.
Any help would be appreciated, thank you
$f(n)$ is completely multiplicative, so for $n=p_1^{a_1}\cdots p_r^{a_r}$ $$g(n):=\sum_{d\mid n}f(d)=\sum_{\substack{b_1,\ldots,b_r\\0\leqslant b_k\leqslant a_k}}f(p_1)^{b_1}\cdots f(p_r)^{b_r}=\prod_{k=1}^r\sum_{b=0}^{a_k}f(p_k)^b.$$ The inner sum is a geometric one, but for our $f(p_k)\in\{-1,0,1\}$ it's easier just to do a bunch of case distinctions. The result is as follows. If there is (at least one) $k$ such that $a_k$ is odd and $f(p_k)=-1$, then $g(n)=0$. Otherwise $g(n)=\prod_{k\ :\ f(p_k)=1}(a_k+1)$.