Equation with Euler's Totient Function

We will prove this by cases

Case 1: $n=ab$ for $1 < a < b < n$ , $gcd(a,b)=1$ . Then $\phi(n)=\phi(a)\phi(b)=100$ , so in particular $\phi(a) | 100$ . Thus, we must have $\phi(a)=1, 2, 4, 5,$ or $10$ . Since $5$ and $25$ are odd and not $1$ , we cannot have $\phi(a)=4, 5$ . If $\phi(a)=1$ , then we must have $a=2$ , and $b$ must be odd (this will be significant later). If $\phi(a)=2$ , then $\phi(b)=50$ , for which there is no solution. Furthermore, $\phi(a)=10$ implies $a=11, 22$ , but then $b$ is not coprime to $a$ .

Case 2: Clearly since $\phi(2^k)=2^{k-1} \ne 100$ , our second case will be $n=p^k$ for odd prime $p$ and $k\geq 1$ . If $k=1$ , we have $n=101$ . If $k=3$ , then we have $n=125$ . All other $k$ do not yield a solution.

Therefore, the values for $n$ are $n =101, 125, 202, 250$