An arithmetic function is a real-valued function whose domain is the set of positive integers. Define the convolution product of two arithmetic functions $f$ and $g$ to be the arithmetic function $f \star g$, where $$(f \star g)(n) = \sum_{ij = n}f(i)g(j), \text{ and } f^{\star k} = f \star f \star \cdots \star f \text{ } (k \text{ times}).$$ We say that two arithmetic functions $f$ and $g$ are dependent if there exists a nontrivial polynomial of two variables $P(x,y) = \sum_{i,j} a_{ij}x^iy^j$ with real coefficients such that $$P(f,g) = \sum_{i,j}a_{ij}f^{\star i} \star g^{\star j} = 0$$ and say that they are independent if they are not dependent. Let $p$ and $q$ be two distinct primes and set $$f_1(n) = \begin{cases}1 \quad \text{if } n = p,\\0 \quad \text{otherwise}; \end{cases} \qquad \qquad f_2(n) = \begin{cases}1 \quad \text{if } n = q,\\0 \quad \text{otherwise}.\end{cases}$$ Prove that $f_1$ and $f_2$ are independent.
My book says that $$f_{1}^{\star i}\star f_{2}^{\star j}(n) = \begin{cases}1 \,\, \text{if} \,\, n=p^{i}q^{j} \\ 0 \,\, \text{otherwise} \end{cases}$$ and the result follows from that, but how do they get that and how does the result follow from that?
Let us write the convolution $$f_1\star f_1 (n) = \sum_{i\cdot j =n}f_1(i)f_1(j).$$
The function $f_1(i)\ne 0\iff i=p$, hence the only way to get a nonzero product inside the sum above is to put $i=j=p$. This implies that $$f_1^{\star 2}(n)=\begin{cases}1,&n=p^2,\\0,&n\ne p^2.\end{cases}$$
By using the same technique, you will get that $$f_1^{\star k}(n)=\begin{cases}1,&n=p^k,\\0,&n\ne p^k.\end{cases}$$
Can you now apply the same technique to $f_1^{\star k}\star f_2^{\star p}$?