5. Let $N=2019^{2}-1$. How many positive factors of $N^{2}$ do not divide $N$ ?
Answer: 157
Solution: Note that $N=2^{3} \cdot 5 \cdot 101 \cdot 1009$ and so $N$ has $4 \cdot 2 \cdot 2 \cdot 2=32$ factors. On the other hand, $N^{2}=2^{6} \cdot 5^{2} \cdot 101^{2} \cdot 1009^{2}$ and so $N^{2}$ has $7 \cdot 3 \cdot 3 \cdot 3=189$ factors. Hence, $189-32=157$ of these factors of $N^{2}$ do not divide $N$.