Problem (Algebra):
For each positive integer $k$, let $S_{k}$ denote the infinite arithmetic sequence of integers with first term $k$ and common difference $k^{2}$. For example, $S_{3}$ is the sequence $3,12,21, \ldots$ Compute the sum of all $k$ such that 306 is an element of $S_{k}$.

Solution:
If 306 is an element of $S_{k}$, then there exists an integer $m \geq 0$ such that $306=k+m k^{2}$. Thus $k \mid 306$ and $k^{2} \mid 306-k$. The second relation can be rewritten as $k \mid 306 / k-1$, which implies that $k \leq \sqrt{306}$ unless $k=306$. The prime factorization of 306 is $2 \cdot 3^{2} \cdot 17$, so the set of factors of 306 less than $\sqrt{306}$ is $\{1,2,3,6,9,17\}$. Check each in turn:

$$
\begin{aligned}
306-1 & =305, & & 1^{2} \mid 305 \\
306-2 & =304, & & 2^{2} \mid 304 \\
306-3 & =303, & & 3^{2} \nmid 303 \\
306-6 & =300, & & 6^{2} \nmid 300 \\
306-9 & =297, & & 9^{2} \nmid 297 \\
306-17 & =289, & & 17^{2} \mid 289 .
\end{aligned}
$$

Thus the set of possible $k$ is $\{1,2,17,306\}$, and the sum is $1+2+17+306=\mathbf{3 2 6}$.

Final answer: 326