Sum of Euler's totient, Möbius and divisor functions Find all the numbers $n$, so that 
$$n=\varphi(n)+\mu(n)+\tau(n),$$
Where $\varphi$ is the Euler's totient function, $\mu$ is the Möbius function and $\tau$ is the sum of positive divisors function.
I could use some help on how to approach this problem.
 A: Here I assume that $n>1$.
Well if $\tau(n)$ is the sum of positive divisors of $n$, then, because $1$ and $n$ always divide $n$ then you have :
$$\tau(n)\geq n+1 $$
Furthermore $\mu(n)\geq -1$ and $\varphi(n)\geq 1$, so you get :
$$\varphi(n)+\mu(n)+\tau(n)\geq n+1>n $$
So in this case it cannot be true and for $n=1$ well it depends a bit of your definitions but I think $\varphi(1)=\mu(1)=\tau(1)=1$ is a good choice so that you cannot have the equality in this case neither. 
Maybe I did not quite understand what you mean by the sum of positive divisors ($\tau$ usually refers to the number of divisors and $\sigma$ the sum fo divisors). The exercice would be maybe more interesting if we had $\tau(n)$ is the number of divisors. Let's try this ? For $n=1$ you can't have the equality. Assume $n=p$ is prime. Then you have :
$$\varphi(p)=p-1\text{, }\mu(p)=-1\text{ and } \tau(p)=2 $$
So you have $p=\varphi(p)+\mu(p)+\tau(p)$ in this case. Maybe if we try $n=p^k$ where $p$ is prime and $k>1$ :
$$\varphi(n)=p^{k-1}(p-1)\text{, }\mu(n)=0\text{ and } \tau(p)=k+1 $$
Then you have :
$$p^{k-1}(p-1)+k+1=p^k\Leftrightarrow p^k-p^{k-1}+k+1=p^k\Leftrightarrow k=p^{k-1}-1 $$
I think that this last equation cannot happen if $p>3$ because $k>1$ implies that (if this equation is true) : 
$$k\geq p-1 $$
But a little analysis show that within this range you cannot have the equation. 
If $p=2$ then $k=2^{k-1}-1$ admits only one case of equality that is $k=3$, otherwise for $k$ even this clearly can't happen and $k$ odd $>3$, $2^{k-1}>>k$. 
If $p=3$ then $k=3^{k-1}-1$ admits only one case of equality that is $k=2$, and if $k>3$, then  $3^{k-1}>>k$.We thus see that within non-trivial prime powers the equality holds for prime numbers and $8$ and $9$. 
Finally suppose first your $n$ is quadratfrei (squarefree) then :
$$n=p_1p_2...p_r $$
with $p_1$,...,$p_r$ being distinct primes. 
$$\varphi(n)=n(1-\frac{1}{p_1})...(1-\frac{1}{p_r})\text{, }\mu(n)=(-1)^r\text{ and } \tau(p)=2^r$$
Let us rewrite :
$$\varphi(n)=n+\sum_{k=1}^r\sum_{A=\{i_1,...,i_k\}\subseteq \{1,...,r\}}\frac{(-1)^kn}{p_{i_1}...p_{i_k}} $$
$$\tau(n)=1+\sum_{k=1}^r\sum_{A=\{i_1,...,i_k\}\subseteq \{1,...,r\}}1$$
So we finally get :
$$\varphi(n)+\mu(n)+\tau(n)=n+1+(-1)^r+\sum_{k=1}^r\sum_{A=\{i_1,...,i_k\}\subseteq \{1,...,r\}}\frac{(-1)^kn}{p_{i_1}...p_{i_k}}+1$$
I am not sure if this is the right way to do this when $n$ is divisible by multiple primes (actually you can use that those three functions are multiplicative maybe it is more efficient). I won't finish this question (a lot of calculus) but if you use Magma (see here for a free session http://magma.maths.usyd.edu.au/calc/) then put this code in the calculator :
N:=10000;
for n:=1 to N do
Q:=Factorization(n);
K:=EulerPhi(Q)+NumberOfDivisors(Q)+MoebiusMu(Q);
if K eq n then if not(IsPrime(n)) then print "------"; n;end if; end if;
end for;
It simply returns every integer below $N$ which is not prime but still verifies the equality.
