I have a set of independent random variables $A_{1}, A_{2}, ... , A_{n}, n \geq 7$ with
$ \forall i \in[1, n], \ \mathbb{E}(A_{i}) = 0 \ \& \text{ Var}(A_{i}) = 1$
I have two other random variables, B and C where
$B = 5A_{p} + 7A_{p-3}, p\geq 4 $ and $C = 13A_{p+7m} - 6A_{p-3+4m}, m \geq 0$
I am attempting to determine Cov(B, C)
Everything I have attempted so far makes me believe the answer is $0$ but I feel this can't be right.
Note that for integers $p, m$ satisfying the given criteria, the set $\{p, p-3, p+7m, p-3+4m\}$ are all distinct unless $m = 0$, since $p = p-3$ is impossible; $p = p+7m$ or $p-3 = p-3 + 4m$ implies $m = 0$; $p = p-3 + 4m$ or $p-3 = p+7m$ implies $m$ is not an integer; and $p+7m = p-3 + 4m$ implies $m$ is negative.
If $m = 0$, then $$\begin{align} \operatorname{Cov}[B,C] &= \operatorname{Cov}[5A_p + 7A_{p-3}, 13A_p - 6A_{p-3}] \\ &= 5(13) \operatorname{Cov}[A_p, A_p] - 5(6)\operatorname{Cov}[A_p, A_{p-3}] + 7(13)\operatorname{Cov}[A_{p-3}, A_p] - 7(6)\operatorname{Cov}[A_{p-3},A_{p-3}] \\ &= 65 \operatorname{Var}[A_p] - 42\operatorname{Var}[A_{p-3}] \\ &= 23. \end{align}$$
If $m \ne 0$, then it is obvious that $\operatorname{Cov}[B,C] = 0$.