## Task 4 - 200934

Determine all pairs $(a ; b)$ of natural numbers $a, b$ with $a > b$, for which the following statements (1), (2), (3) are true!

(1) The number $a$ is (in decimal notation) two-digit, the number $b$ as well.

(2) If the digits of $a$ are swapped, one obtains $b$.

(3) Subtracting $b^2$ from $a^2$ results in a square number.
According to (1) and (2), there exist positive digits $c>d$ with $c, d \in\{1,2, \ldots, 9\}$, such that $a=10 c+d$ and $b=10 d+c$ hold. Then,

$$
a^{2}-b^{2}=\left(100 c^{2}+20 c d+d^{2}\right)-\left(100 d^{2}+20 d c+c^{2}\right)=99\left(c^{2}-d^{2}\right)
$$

is a perfect square, and thus, due to $9=3^{2}$, also $11\left(c^{2}-d^{2}\right)=11(c-d)(c+d)$. Since 11 is a prime number, at least one of the factors $(c-d)$ or $(c+d)$ must also be divisible by 11, so that in the prime factorization of the product, the prime number 11 does not appear only once.

However, since $0<c-d \leq 9-1=8<11$, it must be that $c+d$ is divisible by 11. Given $0<c+d<10+10=20<22$, it follows that $c+d=11$ must hold. Then, $11 \cdot(c-d) \cdot 11=11^{2} \cdot(c-d)$, so $c-d$ must also be a perfect square. Given $0<c-d \leq 9-1=8$, only 1 and 4 are possible as perfect squares, leading to the following two cases:

1. Case: $c-d=1$. Then, due to $c+d=11$, we have $c=6$ and $d=5$. This results in $a=65$ and $b=56$. Indeed, $a^{2}-b^{2}=65^{2}-56^{2}=4225-3136=1089=33^{2}$ is a perfect square, so $(a, b)=(65,56)$ is a solution.
2. Case: $c-d=4$. Then, due to $c+d=11$, we would have $2 c=(c+d)+(c-d)=15$, making the digit $c$ not a natural number, which is a contradiction.

Therefore, there is no other solution besides the one given.