I want to prove that $$\varphi(n)+n\mid \sigma(n)$$ is impossible , if $\ \omega(n)=2\ $ , in other words , $\ n\ $ has exactly two distinct prime factors.
$\ \varphi(n)\ $ is the totient function and $\ \sigma(n)\ $ the divisor-sum-function.
Because of $\ \omega(n)=2\ $ , there are positive integers $\ k,m\ $ and primes $\ p,q\ $ with $\ p<q\ $ , such that $\ n=p^k\cdot q^m\ $
The desired divisibility is $$p^{k-1}q^{m-1}(p-1)(q-1)+p^kq^m \mid\frac{p^{k+1}-1}{p-1}\cdot \frac{q^{m+1}-1}{q-1}$$ so $$p^{k-1}q^{m-1}(2pq-p-q+1)\mid \frac{p^{k+1}-1}{p-1}\cdot \frac{q^{m+1}-1}{q-1}$$
We can conclude $$p^{k-1}\mid q^{m+1}-1$$ and $$q^{m-1}\mid p^{k+1}-1$$ but what to do next ? In particular, how to deal with the factor $\ 2pq-p-q+1\ $ ?