Let $\mu\left(n\right)$ be the Möbius function. Let $\phi\left(n\right)$ be Euler's totient function. Let $\sigma\left(n\right)$ be the sum of divisors and $\tau\left(n\right)$ be the number of divisors functions. I am curious to know whether or not the system:
has at most one solution.
Motivation: I remember a number theory assignment I had where we were given particular values for each of these functions and asked to recover the original number. I can't for the life of me remember how (or if) I managed to solve this problem. I tried to work out a general proof, but couldn't. I also wrote a loop in maple to check for counterexamples, but haven't found any yet. I feel like this is something I should know, but probably have forgotten the relevant facts to approaching this problem.