[Algebra]
Problem: For digits $A, B$, and $C,(\underline{A} \underline{B})^{2}+(\underline{A} \underline{C})^{2}=1313$. Compute $A+B+C$.

Solution: Because $10 A \leq \underline{A} \underline{B}<10(A+1), 200 A^{2}<(\underline{A} \underline{B})^{2}+(\underline{A} \underline{C})^{2}<200(A+1)^{2}$. So $200 A^{2}<$ $1313<200(A+1)^{2}$, and $A=2$. Note that $B$ and $C$ must have opposite parity, so without loss of generality, assume that $B$ is even. Consider the numbers modulo 10: for any integer $n, n^{2} \equiv 0,1,4,5,6$, or $9 \bmod 10$. The only combination whose sum is congruent to $3 \bmod 10$ is $4+9$. So $B=2$ or 8 and $C=3$ or 7 . Checking cases shows that $28^{2}+23^{2}=1313$, so $B=8, C=3$, and $A+B+C=\mathbf{1 3}$.
Rewrite $1313=13 \cdot 101=\left(3^{2}+2^{2}\right)\left(10^{2}+1^{2}\right)$. The two-square identity states:

$$
\begin{aligned}
\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right) & =(a x+b y)^{2}+(a y-b x)^{2} \\
& =(a y+b x)^{2}+(a x-b y)^{2}
\end{aligned}
$$

Therefore

$$
\begin{aligned}
1313=(30+2)^{2}+(3-20)^{2} & =32^{2}+17^{2} \\
& =(3+20)^{2}+(30-2)^{2}=23^{2}+28^{2}
\end{aligned}
$$



Hence $A=2, B=3, C=8$, and $A+B+C=\mathbf{1 3}$.

Final answer: 13