## Task A-2.6.

Let $a$ and $b$ be natural numbers such that $a < b$ and let $S = \{a^2, a^2 + 1, a^2 + 2, \ldots, b^2\}$. Determine all pairs of numbers $a$ and $b$ for which exactly $1\%$ of the elements in the set $S$ are squares of natural numbers.
## Solution.

The set $S$ contains a total of $b^{2}-a^{2}+1$ numbers. 1 point

The perfect squares in the set $S$ are the numbers $a^{2},(a+1)^{2}, \ldots, b^{2}$, and there are a total of $b-a+1$ of them. 2 points

Thus, we are looking for pairs of natural numbers $(a, b)$ such that $\frac{b^{2}-a^{2}+1}{100}=b-a+1$.

Let's rewrite this equation in a different form:

$$
\begin{gathered}
b^{2}-a^{2}+1=100(b-a+1) \\
(b-a)(b+a)=100(b-a)+99 \\
(b-a)(b+a-100)=99
\end{gathered}
$$

Since $b > a$, the factors $b-a$ and $b+a-100$ must be positive divisors of 99. They are $1,3,9,11,33,99$. Therefore, it must be that

$$
(b-a, b+a-100) \in\{(1,99),(3,33),(9,11),(11,9),(33,3),(99,1)\}
$$

Each of these cases is a system of equations in $a$ and $b$ that leads to a solution. All the required pairs of numbers $(a, b)$ are

$$
(1,100),(35,68),(49,60),(51,60),(65,68),(99,100)
$$