3. Find the number of natural numbers $k$, not exceeding 445000, such that $k^{2}-1$ is divisible by 445. Answer: 4000.
Solution. Factoring the dividend and divisor, we get the condition $(k-1)(k+1):(5 \cdot 89)$. This means that one of the numbers $(k+1)$ or $(k-1)$ is divisible by 89. Let's consider two cases.

a) $(k+1): 89$, i.e., $k=89 p+88, p \in \mathrm{Z}$. Then we get $(89 p+87)(89 p+89):(5 \cdot 89) \Leftrightarrow(89 p+87)(p+1): 5$. The first factor is divisible by 5 when $p=5 q+2, q \in \mathrm{Z}$, and the second when $p=5 q+4, q \in \mathrm{Z}$, from which we obtain that $k=445 q+276, k=445 q+444, q \in Z$.

b) $(k-1): 89$, i.e., $k=89 p+1, p \in \mathrm{Z}$. Then we get $89 p(89 p+2):(5 \cdot 89) \Leftrightarrow(89 p+2) p: 5$. The first factor is divisible by 5 when $p=5 q+2, q \in \mathrm{Z}$, and the second when $p=5 q, q \in \mathrm{Z}$, from which we obtain that $k=445 q+179$, $k=445 q+1, q \in \mathrm{Z}$.

Thus, the numbers that satisfy the condition of the problem are those that give remainders $276, 444, 179, 1$ when divided by 445, meaning that every 4 out of 445 consecutive numbers fit. Since $445000=445 \cdot 1000$, we get $4 \cdot 1000=4000$ numbers.