13. It is given that $k$ is a positive integer not exceeding 99. There are no natural numbers $x$ and $y$ such that $x^{2}-k y^{2}=8$. Find the difference between the maximum and minimum possible values of $k$.
13. Answer: 96
Note that $3^{2}-1 \times 1^{2}=8$ and $4^{2}-2 \times 2^{2}=8$. Suppose $k=0(\bmod 3)$. Note that $a^{2}=0,1(\bmod 3)$ for all natural numbers $a$. Thus, $x^{2}-k y^{2}=0,1-0(\bmod 3)=0,1(\bmod 3)$ but $8=2(\bmod 3)$. Hence, there are no natural numbers $x$ and $y$ such that $x^{2}-k y^{2}=8$ if $k$ is a multiple of 3 . Therefore, $\max \{k\}-\min \{k\}=99-3=96$.