Estonian Mathematical Olympiad

Let $a$ be a composite number such that $a+1$ is composite, too. Then $a^2$ is usual since each of its prime divisors is a prime divisor of $a$ and therefore less than $a$. We show that $a^2-1$ is usual, too. To this end, observe that $a^2-1 = (a-1)(a+1)$. As $a+1$ is composite, all prime divisors of $a+1$ are less than $a+1$. But as the difference of $a+1$ and its arbitrary prime divisor is also divisible by this divisor, the prime divisors of $a+1$ cannot be larger than $a-1$. Thus no prime divisor of $a^2-1$ is larger than $a-1$, meaning that the squares of prime divisors of $a^2-1$ do not exceed $(a-1)^2$. As $(a^2-1) - (a-1)^2 = (a-1)((a+1) - (a-1)) = 2(a-1) > 0$, squares of prime divisors of $a^2-1$ are less than $a^2-1$, implying that $a^2-1$ is usual. Consequently, $n = a^2-1$ is a suitable example.
As there exist arbitrarily long sequences of consecutive composite numbers, one can choose two consecutive composite numbers in infinitely many ways. Hence there exist infinitely many integers with the desired property. It remains to notice that $a = 21$ implies $a^2 + 1 = 442 = 2 \cdot 13 \cdot 17$, whereas $2^2 < 13^2 < 17^2 = 289 < 442$. Hence (440, 441, 442) is a triple of consecutive positive integers, all of which are usual.

Consider the so-called Pell's equation $x^2 - 2y^2 = 1$. It has the trivial solution $(x_0, y_0) = (1, 0)$, and whenever $(x_{k-1}, y_{k-1})$ is a solution, $(x_k, y_k) = (3x_{k-1} + 4y_{k-1}, 2x_{k-1} + 3y_{k-1})$ is a solution, too.
As $x_0 < x_1 < x_2 < \dots$ and $y_0 < y_1 < y_2 < \dots$, all these solutions are distinct. Note that $3 \mid y_0$ and, whenever $3 \mid y_{2k}$, also $3 \mid 3x_{2k} + 4y_{2k} = x_{2k+1}$, and whenever $3 \mid x_{2k+1}$, also $3 \mid 2x_{2k+1} + 3y_{2k+1} = y_{2k+2}$. Hence $3 \mid x_{2k+1}$ for every natural number $k$.
Let $k$ be any positive integer. We show that $n = 2y_{2k+1}^2$ meets the conditions. To this end, let $p$ be any prime divisor of $n$. If $p > 2$ then $p \mid y_{2k+1}^2$, implying $p \mid y_{2k+1}$ which gives $p \le y_{2k+1}$. But if $p = 2$ then $p = y_1 < y_{2k+1}$ still. Thus $p^2 \le y_{2k+1}^2 < 2y_{2k+1}^2 = n$ in any case. Let $p$ now be a prime divisor of $n+1 = x_{2k+1}^2$. Then also $p \mid x_{2k+1}$. We previously showed that $3 \mid x_{2k+1}$. As $3 = x_1 < x_{2k+1}$, the number $x_{2k+1}$ is composite, implying that $p < x_{2k+1}$. Thus $p^2 < x_{2k+1}^2 = n+1$. Consequently, $n$ and $n+1$ are both usual. This solves part (a) of the problem since the parameter $k$ can be chosen in infinitely many ways which all produce different solutions.
To solve part (b), it suffices to note that if $k = 1$ then $n = 2 \cdot 70^2 = 9800$, $n+1 = 99^2 = 9801$ and $n+2 = 9802 = 2 \cdot 13^2 \cdot 29$. Moreover, observe that $2^2 < 13^2 < 29^2 = 841 < 9802$. Hence 9800, 9801 and 9802 are three consecutive positive integers which are all usual.