## Task 3 - 330813

Sebastian is considering a three-digit natural number and notices:

(1) If you place the digit 5 in front of this three-digit number, the resulting four-digit number is a perfect square.

(2) If you append the digit 1 to the (original three-digit) number, the resulting four-digit number is also a perfect square.

Prove that there is exactly one three-digit number that satisfies conditions (1) and (2); determine this number!
I. If the conditions (1), (2) are satisfied with a three-digit natural number $z$, then:

Since $z$ is a three-digit number, it follows that $100 \leq z \leq 999$. For the square number $n^{2}$ resulting from (1), it follows that $5100 \leq n^{2} \leq 5999$.

From this, as well as from $71^{2}=5041<5100$ and $5999<6084=78^{2}$, it follows that $71<n<78$. The squares of the numbers $n=72,73,74,75,76,77$ are $n^{2}=5184,5329,5476,5625,5776,5929$.

Therefore, $z$ can only be one of the numbers 184, 329, 476, 625, 776, 929; by appending the digit 1, the numbers 1841, 3291, 4761, 6251, 7761, 9291 are formed.

Of these, only 4761 is a square number, as can be confirmed, for example, by

$$
\begin{array}{rll}
42^{2}=1764<1841<1849=43^{2} & ; \quad 57^{2}=3249<3291<3364=58^{2} \\
69^{2}=4761 & ; \quad 79^{2}=6241<6251<6400=80^{2} \\
88^{2}=7744<7761<7921=89^{2} & ; \quad 96^{2}=9216<9291<9409=97^{2}
\end{array}
$$

Thus, only the number $z=476$ can satisfy the conditions of the problem.

II. It satisfies these conditions; it is a three-digit number, and 5476 as well as 4761 are square numbers.

Thus, the required proof has been provided and the desired number has been determined; it is 476.