2. Find all pairs of natural numbers $a$ and $b$, for which $a-b=101$ and $ab$ is a perfect square.
II/2. 1. way. Let $d$ be the greatest common divisor of the numbers $a$ and $b$. Thus, $a = d m$ and $b = d n$, where $m$ and $n$ are coprime natural numbers. Since $a b = d^2 m n$ is a perfect square and $m$ and $n$ are coprime, $m$ and $n$ must also be perfect squares. Write $m = x^2$ and $n = y^2$, where $x$ and $y$ are natural numbers. It follows that $101 = a - b = d(x + y)(x - y)$. Since 101 is a prime number and $x + y \geq 2$, it must be that $d = x - y = 1$ and $x + y = 101$. From these, we calculate $x = 51$ and $y = 50$, so $a = 51^2 = 2601$ and $b = 50^2 = 2500$.

2. way. Let $d$ be the greatest common divisor of the numbers $a$ and $b$. Since $d$ divides 101 and 101 is a prime number, $d$ can only be 1 or 101.

If $d = 1$, then $a$ and $b$ are coprime numbers. Since $a b$ is a perfect square, it follows that $a$ and $b$ are also perfect squares. Write $a = x^2$ and $b = y^2$, where $x$ and $y$ are natural numbers. Then $101 = a - b = x^2 - y^2 = (x + y)(x - y)$, from which it follows that $x - y = 1$ and $x + y = 101$, since 101 is a prime number and $x + y \geq 2$. From these two equations, we easily calculate $x = 51$ and $y = 50$, so $a = 51^2 = 2601$ and $b = 50^2 = 2500$.

If $d = 101$, we can write $a = 101 m$ and $b = 101 n$, where $m$ and $n$ are coprime natural numbers. Since $a b = 101^2 m n$ is a perfect square and $m$ and $n$ are coprime, it follows that $m$ and $n$ are also perfect squares. Write $m = u^2$ and $n = v^2$, where $u$ and $v$ are natural numbers. Then $101 = a - b = 101(m - n) = 101(u^2 - v^2) = 101(u + v)(u - v)$, which is a contradiction, since $u + v \geq 2$.

The only solution to the problem is thus the pair $a = 2601$ and $b = 2500$.

3. way. Write $a = b + 101$ and calculate $a b = b(b + 101) = n^2$ for some natural number $n$. From this, we express $n^2 - b^2 = (n - b)(n + b) = 101 b$. Since 101 is a prime number, $101 \mid (n - b)$ or $101 \mid (n + b)$. We consider both possibilities.

If $101 \mid (n - b)$, we can write $n - b = 101 k$ for some natural number $k$. From the equation $(n - b)(n + b) = 101 b$, it follows that $k n = b(1 - k)$, which is a contradiction, since the left side of the equation is strictly positive, while the right side is less than or equal to zero.

If $101 \mid (n + b)$, we can write $n + b = 101 l$ for some natural number $l$. From the equation $(n - b)(n + b) = 101 b$, it follows that $l(101 l - 2 b) = b$, or $101 l^2 = b(2 l + 1)$, from which we express $b = \frac{101 l^2}{2 l + 1}$, which must be a natural number. Therefore, $2 l + 1$ divides $101 l^2$. Since 101 is a prime number, it must be that $(2 l + 1) \mid 101$, so we can write $101 = (2 l + 1) c$, where $c \in \mathbb{N}$. Since $2 l + 1 > 1$, it must be that $c = 1$ and $2 l + 1 = 101$. From this, we calculate $l = 50$ and $b = 50^2 = 2500$. Then we calculate $a = 2601$.

1. way:

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-11.jpg?height=51&width=1631&top_left_y=937&top_left_x=218)

Conclusion that $m$ and $n$ are perfect squares .......................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-11.jpg?height=51&width=1633&top_left_y=1037&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-11.jpg?height=51&width=1633&top_left_y=1085&top_left_x=220)

Calculation of the solution $a = 51^2 = 2601$ and $b = 50^2 = 2500 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point

2. way:

Introduction of the greatest common divisor ..................................................................

Conclusion that in the case $d = 1$, the numbers $a$ and $b$ are perfect squares ........... 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-11.jpg?height=52&width=1631&top_left_y=1339&top_left_x=218)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-11.jpg?height=45&width=1631&top_left_y=1391&top_left_x=218)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-11.jpg?height=48&width=1631&top_left_y=1438&top_left_x=218)

Conclusion that in the case $d = 101$, the numbers $m$ and $n$ are perfect squares . ........ 1 point

Conclusion that in the case $d = 101$ we do not get any solution ........................ 1 point

3. way:

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-11.jpg?height=48&width=1627&top_left_y=1638&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-11.jpg?height=51&width=1633&top_left_y=1688&top_left_x=220)

Correct handling of the case when $101 \mid (n - b)$....................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-11.jpg?height=57&width=1633&top_left_y=1785&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-11.jpg?height=49&width=1631&top_left_y=1837&top_left_x=224)