Which is the smallest pair or triplet of consecutive natural numbers, each of which can be written as the sum of two squares of natural numbers? Show that there is no quadruplet of consecutive natural numbers, each of which can be written as the sum of two squares of natural numbers.
Solution. I. It is advisable to regularly list the numbers that can be written as the sum of the squares of two natural numbers, so that we can select the smallest pair, or three consecutive numbers, from among them. For example, let's write the sum of a square number with its predecessors and double it:

| 1 | 4 | 9 | 16 | 25 |
| ---: | ---: | ---: | ---: | :--- |
| 2 | 5 | 10 | 17 | $26 \ldots$ |
|  | 8 | 13 | 20 | $29 \ldots$ |
|  |  | 18 | 25 | $34 \ldots$ |
|  |  |  | 32 | $41 \ldots$ |
|  |  |  |  | $50 \ldots$ |

In the written part of the table, we find the pair 17, 18. Since the smallest number in the next column is also 26, and the numbers in the subsequent columns are even larger, this is the smallest suitable pair.

To find the triplets, it is better to arrange the obtained sums (remaining below 100 for now) in ascending order, in the pattern of the "two-input" square number table in the I. division textbook, writing the numbers whose tens are $0,1,2, \ldots, 9$ in consecutive rows, and the numbers whose units are $0,1,2, \ldots, 9$ in consecutive columns. For example, the cell common to the "0" row and the "5" column corresponds to the number 5, and we denote the representation $1^{2}+2^{2}$ briefly as $1 ; 2$.

|  | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 0 |  |  | $1 ; 1$ |  |  | $1 ; 2$ |  |  | $2 ; 2$ |  |
| 1 | $1 ; 3$ |  |  | $2 ; 3$ |  |  |  | $1 ; 4$ | $3 ; 3$ |  |
| 2 | $2 ; 4$ |  |  |  |  | $3 ; 4$ | $1 ; 5$ |  |  | $2 ; 5$ |
| 3 |  |  | $4 ; 4$ |  | $3 ; 5$ |  |  | $1 ; 6$ |  |  |
| 4 | $2 ; 6$ | $4 ; 5$ |  |  |  | $3 ; 6$ |  |  |  |  |
| 5 | $1 ; 7$ |  | $4 ; 6$ | $2 ; 7$ |  |  |  |  | $3 ; 7$ |  |
|  | $5 ; 5$ |  |  |  |  |  |  |  |  |  |
| 6 |  | $5 ; 6$ |  |  |  | $1 ; 8$ |  |  | $2 ; 8$ |  |
|  |  |  |  |  |  | $4 ; 7$ |  |  |  |  |
| 7 |  |  | $6 ; 6$ | $8 ; 8$ | $5 ; 7$ |  |  |  |  |  |
| 8 | $4 ; 8$ |  | $1 ; 9$ |  |  | $2 ; 9$ |  |  |  | $5 ; 8$ |
| 9 | $3 ; 9$ |  |  |  |  |  |  | $4 ; 9$ | $7 ; 7$ |  |

By writing the decompositions of our previous sums in this table column by column, the first suitable triplet is 72, 73, 74.

II. Among four consecutive integers, two are even and two are odd. The latter can only be the sum of one even and one odd number. Now, the square of an even number can only be the square of an even base, and the square of an odd number can only be the square of an odd base, because if $n$ is an integer, then $(2 n)^{2}=4 n^{2}=4 k$, even, and $(2 n+1)^{2}=4 n(n+1)+1=4 k+1$, odd. When divided by 4, these give a remainder of 0 or 1, so their sum also gives a remainder of 0, 1, or 2 when divided by 4. Therefore, the remainder of a two-term square sum when divided by 4 cannot be 3, whereas the consecutive odd numbers give a remainder of 1 and 3 when divided by 4. This proves the statement.

Kiss Árpád (Budapest, Bláthy O. Erősáramú Ip. Techn. I. o. t.)

Remarks. 1. Among the found numbers, 18 and 72 are the sum of two equal squares. If we relax the conditions and do not accept such, then the first pair is $25=3^{2}+4^{2}, 26=1^{2}+5^{2}$, and the first triplet is $232=6^{2}+14^{2}, 233=8^{2}+13^{2}$, $234=3^{2}+15^{2}$.

Surányi László (Budapest, V., Szemere u. Ált. Isk. VIII. o. t.)

2. The first number of the suitable triplets is divisible by 8. According to the above, only triplets of the form $8 k, 8 k+1,8 k+2$ and $8 k+4,8 k+5$, $8 k+6$ can be considered. However, the latter is impossible. It is known that if an even number can be expressed as the sum of two integer squares, then its half can also be expressed as such:

$$
2 x=a^{2}+b^{2} \text { implies } \quad x=\left(\frac{a+b}{2}\right)^{2}+\left(\frac{a-b}{2}\right)^{2}
$$

and applying this to $8 k+6$ would result in $4 k+3$ being expressible.

Furthermore, one of the numbers in a suitable triplet is divisible by 3, and therefore by 9. The squares of numbers of the form $3 n, 3 n+1,3 n+2$ are

$$
9 n^{2}, \quad 3\left(3 n^{2}+2 n\right)+1=3 k+1, \quad 3\left(3 n^{2}+4 n+1\right)+1=3 k+1
$$

The sum of two such numbers can only be divisible by 3 if neither term is of the form $3 k+1$, so the representation is of the form $9\left(n_{1}^{2}+n_{2}^{2}\right)=9 k$. Therefore, the smallest number of a suitable triplet is divisible by 8 and either divisible by 9, or 1 or 2 less than a number divisible by 9. It is easy to see that then this number gives a remainder of 0, 8, or 16 when divided by 72.

Lovász László (Budapest, Fazekas M. Gyak. Gimn. I. o. t.)

3. The triplet $16=0^{2}+4^{2}, 17$, 18 does not meet the criteria because 0 is not a natural number (several gave this).

[^0]:    ${ }^{1}$ See, for example, Hajós-Neukomm-Surányi: Mathematical Theorems II. (Tankönyvkiadó, Budapest 1967.) p. 59.