## Task 1 - 230821

Determine all four-digit natural numbers $z$ that satisfy the following conditions:

(1) The two-digit number formed by the first two digits of $z$ in this order is a perfect square.

(2) The number formed by the first and fourth digits of $z$ in this order is also a perfect square.

(3) The number formed by the second and third digits of $z$ in this order is also a perfect square.

Hint: By the first digit, we mean the digit of $z$ that is in the thousands place.
For the first two digits of $z$, only the square numbers 16, 25, 36, 49, 64, 81 are possible due to (1). Of these, the numbers 25 and 49 are excluded due to (3), as there are no two-digit square numbers starting with the digits 5 or 9.

If we start from the remaining numbers 16, 36, 64, and 81, then due to (3), only the digits 4, 4, 9, and 6 can stand at the third position of $z$, and due to (2), only the digits 6, 6, 4, and 1 can stand at the fourth position.

Consequently, only the four numbers 1646, 3646, 6494, and 8161 can have all the required properties.

These numbers have these properties, as (for the first two digits and also for the first and fourth digit) 16, 36, 64, 31 and (for the second and third digits) 64, 64, 49, and 16 are all square numbers.