3. Three students wrote a two-digit number on the board, each of which is a perfect square. It turned out that if you "glue" them into one six-digit number, it is also a square of a natural number. Find all such six-digit numbers.
Solution. Let the two-digit squares be $x^{2}, y^{2}$, and $z^{2}$, then $x, y, z \in\{4,5,6,7,8,9\}$. The six-digit number is denoted by $t^{2}(t>0)$. Then

$$
t^{2}=10000 x^{2}+100 y^{2}+z^{2}=(100 x)^{2}+(10 y)^{2}+z^{2}
$$

Let $t=100 x+k$. Obviously, $k \in \mathbb{N}$, since $t^{2}>(100 x)^{2}$. Then

$$
\begin{aligned}
(100 x)^{2}+(10 y)^{2}+z^{2} & =(100 x+k)^{2} \\
100 y^{2}+z^{2} & =200 k x+k^{2}
\end{aligned}
$$

Let's consider the cases. If $k \geqslant 11$, then the right side $\geqslant 200 \cdot 11 \cdot 4+11^{2}=8921$, but the left side $\leqslant 8181$. If $k=10$, then the right side is divisible by 100, but the left side is not. Finally, let $k<10$, then $k^{2}<100$, so $100 y^{2}+z^{2}<200 k x+100$. Since $x \geq 4$, we have $200 k x \geq 800 k$, so $100 y^{2}+z^{2}<800 k+100$. The only possible values for $k$ are 1, 2, 3, 4, 5, 6, 7, 8, and 9. We can check each case:

- If $k=1$, then $100 y^{2}+z^{2}=200 x+1$, which is impossible since the left side is even and the right side is odd.
- If $k=2$, then $100 y^{2}+z^{2}=400 x+4$, which is impossible since the left side is not divisible by 4.
- If $k=3$, then $100 y^{2}+z^{2}=600 x+9$, which is impossible since the left side is not divisible by 3.
- If $k=4$, then $100 y^{2}+z^{2}=800 x+16$, which is impossible since the left side is not divisible by 4.
- If $k=5$, then $100 y^{2}+z^{2}=1000 x+25$, which is impossible since the left side is not divisible by 5.
- If $k=6$, then $100 y^{2}+z^{2}=1200 x+36$, which is impossible since the left side is not divisible by 4.
- If $k=7$, then $100 y^{2}+z^{2}=1400 x+49$, which is impossible since the left side is not divisible by 7.
- If $k=8$, then $100 y^{2}+z^{2}=1600 x+64$, which is possible if $x=4$ and $y=8$.
- If $k=9$, then $100 y^{2}+z^{2}=1800 x+81$, which is impossible since the left side is not divisible by 9.

In the case $k=8$, we have $100 y^{2}+z^{2}=1600 x+64$. If $x=4$, then $100 y^{2}+z^{2}=6464$. The only solution is $y=8$ and $z=4$. Therefore, the six-digit number is $166464=408^{2}$ and $646416=804^{2}$.

Answer: $166464=408^{2}$ and $646416=804^{2}$.