Along with the Chinese Remainder theorem, you need to be able to factor your $k$ into prime powers. It is a theorem requiring induction that the only solutions to $ n^2 \equiv n \pmod {p^t} $ for a prime $p$ and $t \geq 1$ are $ n \equiv 0,1 \pmod {p^t}. $ As a result, if your $k$ is divisible by exactly $r$ distinct primes, there are exactly $2^r$ of your fixpoints. For $100,$ note that you got precisely the four numbers from $0$ to $99$ that are $0,1 \pmod 4$ and $0,1 \pmod {25}.$
Let's see, the theorem mentioned is straightforward for odd $p,$ induction on $t.$ Slightly different for $p=2$ but easy enough and still true.
Induction, odd $p$: with $0 \leq j \leq p-1,$ then $\delta = 0,1,$ and $t \geq 1,$ we are solving $ (j p^t + \delta)^2 \equiv j p^t + \delta \pmod {p^{t+1}}, $ $ j^2 p^{2t} + 2 j \delta p^t + \delta^2 \equiv j p^t + \delta \pmod {p^{t+1}}. $ Note $\delta^2 = \delta,$ and $2t \geq t + 1.$ So $ 2 j \delta p^t \equiv j p^t \pmod {p^{t+1}}, $ $ (2 \delta - 1) j p^t \equiv 0 \pmod {p^{t+1}}, $ $ (2 \delta - 1) j \equiv 0 \pmod {p}, $ $ j \equiv 0 \pmod {p}, $ $ j = 0. $
Induction, $p=2$: with $j = 0,1,$ then $\delta = 0,1,$ and $t \geq 1,$ we are solving $ (j 2^t + \delta)^2 \equiv j 2^t + \delta \pmod {2^{t+1}}, $ $ j^2 2^{2t} + 2 j \delta 2^t + \delta^2 \equiv j 2^t + \delta \pmod {2^{t+1}}. $ Note $\delta^2 = \delta,$ and $2t \geq t + 1.$ So $ 2 j \delta 2^t \equiv j 2^t \pmod {2^{t+1}}, $ $ j \delta 2^{t+1} \equiv j 2^t \pmod {2^{t+1}}, $ $ 0 \equiv j 2^t \pmod {2^{t+1}}, $ $ 0 \equiv j \pmod {2}, $ $ j = 0. $