If $\overline{abcd} = (\overline{ab} + \overline{cd})^2$ and only $c$ can be $0$, find the sum of all possible values of $\overline{abcd}$.
$(A) 13850$
$(B) 14051$
$(C) 14742$
$(D) 14851$
$(E) 16051$
First I expanded the RHS to get $$100(a^2+c^2)+b^2+d^2+2(100ac+10ab+10ad+10bc+10cd+bd)=1000a+100b+10c+d$$
Since that didn't seem correct, I plugged in values and it looked like anything above $50^2$ didn't work. Any better ways?
EDIT: I found $99^2$ to work.
Ok, so I went full tryhard mode and literally bashed out everything from $32^2$ to $75^2$. Since the answer choices mean that the last one's in that range, maybe I miscalculated, but I didn't find anything.
Modding by 99, we find that this becomes solving for $$n=n^2 \mod 99$$ $$n(n-1)=0\mod 99$$
Using Chinese remainder theorem, we find the solutions to be $1,45,55,99$. $1$ doesn't work, all other ones work, so the solution is: $45^2+55^2+99^2=14851$