A standard exercise in elementary number theory is to prove that if two multiplicative functions agree on the set of prime powers, then they are identical arithmetic functions.
Consider the following related problem: Suppose one has two arithmetic functions $f$ and $g$ such that they agree on a positive integer $n$, that is, $f(n) = g(n)$. We assume further that $f$ is multiplicative on $\mathbb{N}$, $\displaystyle f(n) = \prod_{p^r \| n} f(p^r)$, but we assume nothing more for $g$. In particular, we do not assume that $g$ is multiplicative.
Question: Under what conditions is it true that $\displaystyle g(n) = \prod_{p^{r} \| n} g(p^r)$, that is, $g$ is pointwise multiplicative on $n$?
Consider two examples. Suppose we have $\varphi(n) = n - 6$, where $\varphi$ is the Euler totient function. The right hand side is decidedly not multiplicative, while the left side is. Note that the equation is valid only for $n = 10$, and one has $\varphi(10) = 4 = 10 - 6 = (2-6)(5-6)$, so $n-6$ is pointwise multiplicative on $n = 10$. However, consider the sum-of-divisors function $\sigma$ and the equation $\sigma(n) = n^2 - 10n + 4$. One solution is $n = 12$, but the right side is not pointwise multiplicative on $n = 12$.