[Number Theory]
Problem: Let $T=13$. Compute the least integer $n>2023$ such that the equation $x^{2}-T x-n=0$ has integer solutions.

Solution: The discriminant of the quadratic, $T^{2}+4 n$, must be a perfect square. Because $T$ and the discriminant have the same parity, and the leading coefficient of the quadratic is 1 , by the quadratic formula, the discriminant being a perfect square is sufficient to guarantee integer solutions. Before knowing $T$, note that $\sqrt{4 \cdot 2024}=$ $\sqrt{8096}$ is slightly less than 90 because $90^{2}=8100$, and the square root must have the same parity as $T$. Because



$T=13$, the square root must be greater than $\sqrt{13^{2}+4 \cdot 2023}=\sqrt{8261}$, which is between 90 and 91 , so the desired square root is 91 . Hence $13^{2}+4 n=91^{2}$, so $n=\mathbf{2 0 2 8}$.

Final answer: 2028