Let's first find the squares between $200$ and $2000.$ They are $15^2,16^2,\cdots,44^2$ because $14^2=196,15^2=225,44^2=1936,$ and $45^2=2025.$ We know $101(A+C)+2B$ must be one of these squares. Let's say $X=A+C$ and $Y=2B.$ Taking this mod $100,$ we see that $X+Y$ must be the residue of a square mod $100.$ With this in mind, the square itself is $100X+(X+Y)$ meaning $X \in \{ 1,2,\cdots,19 \}.$ So, we can find a square's $X$ easily, then find the $Y$ easily as well. This only works if a square's last two digits are greater than of equal to the rest of the number (like for $1234,$ we have $34\ge12$). Notice that $Y$ must be even and no greater than $18.$ Now, we can blitz through the squares! (This part would actually be faster to write than type lol) \begin{align*} 15^2 = 225 &\rightarrow Y = 23 \\ 16^2 = 256 &\rightarrow Y = 54 \\ 17^2 = 289 &\rightarrow Y = 87 \\ 18^2 = 324 &\rightarrow Y = 21 \\ 19^2 = 361 &\rightarrow Y = 58 \\ 20^2 = 400 &\rightarrow \text{come back to this} \\ 21^2 = 441 &\rightarrow Y = 37 \\ 22^2 = 484 &\rightarrow Y = 80 \\ 23^2 = 529 &\rightarrow Y = 24 \\ 24^2 = 576 &\rightarrow Y = 71 \\ 25^2 = 625 &\rightarrow Y = 19 \\ 26^2 = 676 &\rightarrow Y = 70 \\ 27^2 = 729 &\rightarrow Y = 22 \\ 28^2 = 784 &\rightarrow Y = 77 \\ 29^2 = 841 &\rightarrow Y = 33 \\ 30^2 = 900 &\rightarrow \text{come back to this} \\ 31^2 = 961 &\rightarrow Y = 52 \\ 32^2 = 1024 &\rightarrow Y = 14 \rightarrow \text{come back to solve} \\ 33^2 = 1089 &\rightarrow Y = 79 \\ 34^2 = 1156 &\rightarrow Y = 45 \\ 35^2 = 1225 &\rightarrow Y = 13 \\ 36^2 = 1296 &\rightarrow Y = 84 \\ 37^2 = 1369 &\rightarrow Y = 56 \\ 38^2 = 1444 &\rightarrow Y = 30 \\ 39^2 = 1521 &\rightarrow Y = 6 \rightarrow \text{come back to solve} \\ 40^2 = 1600 &\rightarrow \text{come back to this} \\ 41^2 = 1681 &\rightarrow Y = 65 \\ 42^2 = 1764 &\rightarrow Y = 47 \\ 43^2 = 1849 &\rightarrow Y = 31 \\ 44^2 = 1936 &\rightarrow Y = 17 \end{align*}Whew! Ok, we've gotten through most squares. First, let's deal with the "come back to this" cases. These were $400,900,$ and $1600.$ We have $X+Y=A+2B+C \in [0,36]$ since $A,B,C$ are digits, but it's also the residue mod $100,$ so we must have $X+Y=0$ meaning $A=B=C=0$ which is impossible. In conclusion, these are not solutions. Next, let's solve the exciting cases, where we seem to have what we want. These were $1024$ and $1521.$ For $1024,$ we have $Y=14$ meaning $B=7,$ and $X=10$ meaning $A+C=10.$ So, the solutions here are $179,278,377,476,575,674,773,872,$ and $971.$ For $1521,$ we have $Y=6$ meaning $B=3,$ and $X=15$ meaning $A+C=15.$ So, the solutions here are $639,738,837,$ and $936.$ So, here are all the solutions: $179,278,377,476,575,674,773,872,971,639,738,837,936.$ Sidenote: this could be a good AIME problem, where you ask for a palindromic Lusophon number, since $575$ is the only one.

Let's first find the squares between $200$ and $2000.$ They are $15^2,16^2,\cdots,44^2$ because $14^2=196,15^2=225,44^2=1936,$ and $45^2=2025.$ We know $101(A+C)+2B$ must be one of these squares. Let's say $X=A+C$ and $Y=2B.$ Taking this mod $100,$ we see that $X+Y$ must be the residue of a square mod $100.$ With this in mind, the square itself is $100X+(X+Y)$ meaning $X \in \{ 1,2,\cdots,19 \}.$ So, we can find a square's $X$ easily, then find the $Y$ easily as well. This only works if a square's last two digits are greater than of equal to the rest of the number (like for $1234,$ we have $34\ge12$). Notice that $Y$ must be even and no greater than $18.$ Now, we can blitz through the squares! (This part would actually be faster to write than type lol) \begin{align*} 15^2 = 225 &\rightarrow Y = 23 \\ 16^2 = 256 &\rightarrow Y = 54 \\ 17^2 = 289 &\rightarrow Y = 87 \\ 18^2 = 324 &\rightarrow Y = 21 \\ 19^2 = 361 &\rightarrow Y = 58 \\ 20^2 = 400 &\rightarrow \text{come back to this} \\ 21^2 = 441 &\rightarrow Y = 37 \\ 22^2 = 484 &\rightarrow Y = 80 \\ 23^2 = 529 &\rightarrow Y = 24 \\ 24^2 = 576 &\rightarrow Y = 71 \\ 25^2 = 625 &\rightarrow Y = 19 \\ 26^2 = 676 &\rightarrow Y = 70 \\ 27^2 = 729 &\rightarrow Y = 22 \\ 28^2 = 784 &\rightarrow Y = 77 \\ 29^2 = 841 &\rightarrow Y = 33 \\ 30^2 = 900 &\rightarrow \text{come back to this} \\ 31^2 = 961 &\rightarrow Y = 52 \\ 32^2 = 1024 &\rightarrow Y = 14 \rightarrow \text{come back to solve} \\ 33^2 = 1089 &\rightarrow Y = 79 \\ 34^2 = 1156 &\rightarrow Y = 45 \\ 35^2 = 1225 &\rightarrow Y = 13 \\ 36^2 = 1296 &\rightarrow Y = 84 \\ 37^2 = 1369 &\rightarrow Y = 56 \\ 38^2 = 1444 &\rightarrow Y = 30 \\ 39^2 = 1521 &\rightarrow Y = 6 \rightarrow \text{come back to solve} \\ 40^2 = 1600 &\rightarrow \text{come back to this} \\ 41^2 = 1681 &\rightarrow Y = 65 \\ 42^2 = 1764 &\rightarrow Y = 47 \\ 43^2 = 1849 &\rightarrow Y = 31 \\ 44^2 = 1936 &\rightarrow Y = 17 \end{align*}Whew! Ok, we've gotten through most squares. First, let's deal with the "come back to this" cases. These were $400,900,$ and $1600.$ We have $X+Y=A+2B+C \in [0,36]$ since $A,B,C$ are digits, but it's also the residue mod $100,$ so we must have $X+Y=0$ meaning $A=B=C=0$ which is impossible. In conclusion, these are not solutions. Next, let's solve the exciting cases, where we seem to have what we want. These were $1024$ and $1521.$ For $1024,$ we have $Y=14$ meaning $B=7,$ and $X=10$ meaning $A+C=10.$ So, the solutions here are $179,278,377,476,575,674,773,872,$ and $971.$ For $1521,$ we have $Y=6$ meaning $B=3,$ and $X=15$ meaning $A+C=15.$ So, the solutions here are $639,738,837,$ and $936.$ So, here are all the solutions: $179,278,377,476,575,674,773,872,971,639,738,837,936.$ Sidenote: this could be a good AIME problem, where you ask for a palindromic Lusophon number, since $575$ is the only one.