* If the integer $k$ is added to $36, 300, 596$, respectively, the results are the squares of three consecutive terms in an arithmetic sequence, find $k$.
According to the problem, let $36+k=(a-d)^{2}, 300+k=a^{2}, 596+k=(a+d)^{2}$. Thus, $2 a d-d^{2}=264, 2 a d+d^{2}=296 \cdots$ (1). Therefore, $2 d^{2}=32, d=4$. Substituting into (1) yields $a=35$. Hence, $k=(35-4)^{2}-36=925$.