The South African Mathematical Olympiad Third Round

Write $n$ as $a_0+10a_1+100a_2+\dots$, where $a_0,a_1,a_2,\dots$ are the digits of $n$. Then the stated equation is equivalent to $$a_0+10a_1+100a_2+\cdots =a_0+a_1+a_2+\cdots +a_0^2$$ or $$9a_1+99a_2+999a_3+\cdots =a_0^2.$$ The right hand side is at most $9^2=81<99$. Therefore, $a_2,a_3,\dots$ have to be 0 (otherwise, the left hand side would be strictly greater than the right hand side). Hence we obtain $$9a_1=a_0^2.$$ It follows that $a_0$ must be divisible by 3 (since $9a_1$ is), which leaves us with the possibilities $a_0=3$ ($a_1=1$), $a_0=6$ ($a_1=4$) and $a_0=9$ ($a_1=9$). So there are three solutions: 13, 46 and 99.