Let's determine the digits $a, b, c, d$ if the number written in the decimal system with these digits satisfies:

$$
\overline{a b c d}=16\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+a+b
$$

and additionally, $b^{2}-a^{2}=2\left(c^{2}+d^{2}\right)$.
I. solution. Using (2), (1) can be transformed as follows:

$$
\begin{gathered}
10^{3} a+10^{2} b+10 c+d=8 a^{2}+24 b^{2}+a+b \\
999 a+99 b+10 c+d=8 a^{2}+24 b^{2}
\end{gathered}
$$

Here, no term is negative. Therefore, on the left side, keeping only the first term, this side decreases or remains unchanged. The right side, however, increases or remains unchanged if we substitute the largest possible values for \(a\) and \(b\), which is 9. Thus, we get the following inequality:

$$
999 a \leqq (8+24) 9^{2}=2592, \quad \text { from which } \quad a \leqq 2 .
$$

On the other hand, as the leading digit, \(a \geqq 1\). However, \(a=2\) is not possible, because then from (3) we get

$$
10 c+d=24 b^{2}-99 b-1966
$$

and here the right side is negative for every digit \(b\), indeed

$$
24 b^{2}-99 b-1966 \leqq 24 b^{2}-1966 \leqq 24 \cdot 81-1966=-22
$$

Thus, it remains that \(a=1\). A similar solution shows that only \(b=9\) is possible with \(a=1\), because with \(a=1, b=8\) we still get \(10 c+d=-254\), which is impossible. Thus, for \(c, d\) from (2) and (3) we get

$$
c^{2}+d^{2}=40, \quad 10 c+d=62
$$

Of the two solutions to this system, only \(c=6, d=2\) satisfy the condition of being digits. - Therefore, the sought number is 1962; this indeed satisfies both conditions.

Bak Zsuzsanna (Ráckeve, Ady E. Gimn. III. o. t.)

II. solution. According to (2), \(b^{2}-a^{2}\) is even, so either both \(b^{2}\) and \(a^{2}\) are even, or both are odd. Thus, the same holds for \(b\) and \(a\), because the base of an even square number is even, and the base of an odd square number is odd. Therefore, \(a+b\) is even, and thus the right side of (1) is also even, so \(d\) is even. Furthermore, \(c^{2}+d^{2}\) is also even, because it equals half of \(b^{2}-a^{2}\), and \(b^{2}-a^{2}=(b-a)(b+a)\), both factors of which are even, so the product is divisible by 4. Therefore, \(c^{2}\) and thus \(c\) is also even.

Moreover, according to (2), \(b>a\), and the largest possible value of the left side is \(9^{2}-1^{2}=80\), so \(c^{2}+d^{2}\) cannot be greater than 40. Thus, for the pair of digits \(c, d\), only the following 7 value pairs are possible, from which we can easily find the possible pairs of \(a, b\) values based on (2), and thus calculate the value of \(J\) on the right side of (1). (The right side is symmetric in \(c\) and \(d\), so swapping \(c\) and \(d\) is unnecessary.) If the number formed by the digits \(a, b, c, d\) thus used is obtained, then we have a solution. Only the first such trial is successful, the sought number is 1962.

| \(c, d\) | \(b^{2}-a^{2}\) | \(b\) | \(a\) | \(J\) |  |
| :---: | :---: | :---: | :---: | :---: | :--- |
| 6,2 | 80 | 9 | 1 | 1962 | solution |
| 6,0 | 72 | 9 | 3 | 2028 |  |
| 4,4 | 64 | 8 | 0 | does not fit because \(a>0\) |  |
| 4,2 | 40 | 7 | 3 | 1258 |  |
| 4,0 | 32 | 6 | 2 | 904 |  |
|  |  | 9 | 7 | 2352 |  |
| 2,2 | 16 | 5 | 3 | 680 |  |
| 2,0 | 8 | 3 | 1 | 228 |  |

Hanák Péter (Budapest, Fazekas M. Gyak. Gimn. III. o. t.)