3. Find the number of natural numbers $k$, not exceeding 267000, such that $k^{2}-1$ is divisible by 267. Answer: 4000.
Solution. Factoring the dividend and divisor, we get the condition $(k-1)(k+1):(3 \cdot 89)$. 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 $(89 p+87)(89 p+89):(3 \cdot 89) \Leftrightarrow(89 p+87)(p+1): 3$. The first factor is divisible by 3 when $p=3 q, q \in \mathrm{Z}$, and the second when $p=3 q+2, q \in \mathrm{Z}$, from which we obtain that $k=267 q+88, k=267 q+266, q \in \mathrm{Z}$.

b) $(k-1): 89$, i.e., $k=89 p+1, p \in \mathrm{Z}$. Then we get $89 p(89 p+2):(3 \cdot 89) \Leftrightarrow(89 p+2) p: 3$. The first factor is divisible by 3 when $p=3 q+2, q \in \mathrm{Z}$, and the second when $p=3 q, q \in \mathrm{Z}$, from which we obtain that $k=267 q+179$, $k=267 q+1, q \in \mathrm{Z}$.

Thus, the numbers that satisfy the condition of the problem are those that give remainders $88,266,179,1$ when divided by 267, meaning that every 4 out of 267 consecutive numbers fit. Since $267000=267 \cdot 1000$, we get $4 \cdot 1000=4000$ numbers.