Let's check the following equalities, look for regularities in the table, and based on that, write down the next two rows of the table.

$$
\begin{aligned}
3^{2}+4^{2} & =5^{2} \\
10^{2}+11^{2}+12^{2} & =13^{2}+14^{2} \\
21^{2}+22^{2}+23^{2}+24^{2} & =25^{2}+26^{2}+27^{2} \\
36^{2}+37^{2}+38^{2}+39^{2}+40^{2} & =41^{2}+42^{2}+43^{2}+44^{2}
\end{aligned}
$$
The equalities hold, the values on both sides in the subsequent rows are 25, 365, 2030, 7230, respectively.

In each row, the squares of consecutive natural numbers are listed, the number of terms in the subsequent rows is 3, 5, 7, 9, and on the left side, there is always one more term than on the right side. According to this, in the 5th row, 11, and in the 6th row, 13 squares of consecutive natural numbers should be written in such a way that on the left, 6 and 7 terms stand, respectively, and the others on the right side.

The bases of the middle terms in the rows show a recognizable regularity, the numbers 4, 12, 24, 40 also do. The growth among them is 8, 12, 16, each 4 greater than the previous one. If this regularity remains valid, then in the next row, the square of the number 20 greater than 40, 60, stands before the equality sign, and in the following row, the square of the number 24 greater, 84. Writing these as the middle terms, the starting terms of the 5th row are $60-5=55$, and $84-6=78$ for the 6th row. The resulting equalities are indeed correct:

$$
\begin{aligned}
55^{2}+56^{2}+57^{2}+58^{2}+59^{2}+\mathbf{6 0}^{2} & =61^{2}+62^{2}+63^{2}+64^{2}+65^{2} \\
78^{2}+79^{2}+80^{2}+81^{2}+82^{2}+83^{2}+\mathbf{8 4}^{2} & =85^{2}+86^{2}+87^{2}+88^{2}+89^{2}+90^{2}
\end{aligned}
$$

The value of both sides is 19,855, and 45,955, respectively. This satisfies the requirements of the problem.

Jermendy György (Szeghalom, Péter A. g. I. o. t.)

Remarks. 1. The middle $x$ number in the 5th row can also be obtained from the equation

$$
\begin{gathered}
(x-5)^{2}+(x-4)^{2}+(x-3)^{2}+(x-2)^{2}+(x-1)^{2}+x^{2}= \\
\quad=(x+1)^{2}+(x+2)^{2}+(x+3)^{2}+(x+4)^{2}+(x+5)^{2}
\end{gathered}
$$

$x^{2}-60 x=x(x-60)=0$, from which $x_{1}=60$ and $x_{2}=0$. The second root is irrelevant. Similarly, the middle $y$ number in the 6th row from the equation

$$
(y-6)^{2}+(y-5)^{2}+\ldots+y^{2}=(y+1)^{2}+\ldots+(y+6)^{2}
$$

$y_{1}=84$ (another root is again $y_{2}=0$).

Boldizsár Ferenc (Kaposvár, Táncsics M. g. II. o. t.)

2. The starting and ending terms of the subsequent rows show a similar regularity: the increments are 7, 11, 15, and 9, 13, 17, respectively, each step by step 4 greater. From these, we can also obtain the appropriate terms of the subsequent rows.

Several solutions started from the observation that the first base number of each subsequent row is greater than the last base number of the previous row, and between the subsequent rows, 4, 6, 8 natural numbers were missing from the sequence of bases: 6, 7, 8, and 9, then 15, 16, ..., 20, and finally 28, 29, ..., 35. If this regularity continues, then the 5th row starts with $44+(10+1)=55$; and since its ending term is $55+(11-1)=65$, the next row starts with $65+(12+1)=78$ and ends with $78+(13-1)=90$, containing the natural numbers as bases. This is also a correct insight.