The convolution inverse of an arithmetic function

Good, but does an inverse g of f have to exist? A necessary condition is that f(1) 6= 0. Indeed, if g is the inverse of f , then 1 = I(1) = (f ∗ g)(1) = f(1)g(1). We now show that this necessary condition is also sufficient. We assume that f(1) 6= 0 and we try and solve for g. What this means is that we are solving for infinitely many unknowns: g(1), g(2), . . . . From the above, we see that the first unknown g(1) is 1/f(1), so at least that part is solved. Suppose that it is solved for numbers below n, where n ≥ 2. That is, we are supposing that we know the values g(1), g(2), . . . , g(n − 1), and for each number m with 2 ≤ m ≤ n− 1, we have (f ∗ g)(m) = 0. We would like to adjust g so that also (f ∗ g)(n) = 0, but we don’t want to mess up the earlier equations (f ∗ g)(m) = 0. So, the earlier ones are not affected if we do not change the values of g(1), g(2), . . . , g(n− 1) already chosen. To find g(n), we have 0 = (f ∗ g)(n) = ∑