W2 ** Find all non-negative integers $c$ and positive integers $n$, satisfying $\varphi(n)+d(n)=n+c$.


Parse the numbers in $2,3, \cdots, n$. If a number is coprime with $n$, then it is not a divisor of $n$. Only 1 is both a divisor of $n$ and coprime with $n$. Therefore, $\varphi(n)+d(n) \leqslant n+1$, which means $c=0$ or 1.

If $n$ is even and greater than 8, then $n-2, n-4$ are not divisors of $n$, and they are not coprime with $n$, thus
$$
\varphi(n)+d(n) \leqslant n-1 .
$$

If $n$ is an odd composite number and greater than 9, let $p$ be the smallest prime factor of $n$. Then the even numbers $n-p, n-3p$ are not divisors of $n$, and they are not coprime with $n$. Therefore, (1) holds.
When $n$ is 1 or a prime number, $\varphi(n)+d(n)=n+1$;
When $n=4$, the above equation also holds;
When $n=6,8,9$, $\varphi(n)+d(n)=n$.
Thus, the solutions are $(n, c)=(1,1),(4,1),(p, 1),(6,0),(8,0),(9,0)$, where $p$ is a prime number.