Are there any integers $n$ such that $\sigma_0(n) = n$?

Question

Where $\sigma_0(n)$ is the divisor function, are there any integers $n$ such that $\sigma_0(n) = n$? I already know that $n=1, 2$ work, but are there any others?

Answer 1

Hint: No, show that for $p>2$ prime $\sigma _0(p)<p$ and show that for every number $n+1<2^n\leq p^n$ $n>1.$
For a general number, decompose it in prime factors, and use that $\sigma _0 (n\cdot m)=\sigma_0 (n)\cdot\sigma _0(m),$ for $(n,m)=1$