What is the largest natural number $k$ for which there are infinitely many sequences of $k$ consecutive natural numbers, each of which can be written as the sum of two square numbers? (0 is also considered a square number.)
A square number, when divided by 4, leaves a remainder of 0 or 1, so the sum of two square numbers cannot leave a remainder of 3 when divided by 4. Therefore, among four or more consecutive natural numbers, there is one that cannot be expressed as the sum of two square numbers, and thus the largest value of $k$ can be at most 3.

We will show that this largest value is precisely 3: we will provide infinitely many triples of consecutive numbers, all of whose elements can be written as the sum of two square numbers. We will call numbers with this property "beautiful."

It seems natural to examine triples of the form $\left\{m^{2}-1, m^{2}, m^{2}+1\right\}$, since the second and third elements are beautiful $\left(m^{2}+0^{2}, m^{2}+1^{2}\right)$, and thus it suffices to show that $m^{2}-1$ is beautiful for infinitely many natural numbers $m$.

An identity from Euler states that the product of beautiful numbers is also beautiful. Indeed,

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

If $m^{2}-1$ is beautiful, then

$$
\left(m^{2}\right)^{2}-1=\left(m^{2}-1\right)\left(m^{2}+1\right)
$$

and thus, by (1), $\left(m^{2}\right)^{2}-1$ is beautiful as a product of beautiful numbers.

It suffices, therefore, to find a single square number $a^{2}$ such that $a^{2}-1$ is beautiful, because then, according to the above, choosing $a_{0}=a$, $a_{n}=a_{n-1}^{2}$, every element of the sequence $a_{n}^{2}-1$ is beautiful.

Since $8=3^{2}-1=2^{2}+2^{2}$, $a=3$ is suitable. The sequence $3^{2^{n}}-1(n \geqq 1)$ thus obtained has every element beautiful, and thus the triple

$$
\left\{3^{2^{n}}-1,3^{2^{n}}, 3^{2^{n}}+1\right\}
$$

has all its elements expressible as the sum of two square numbers for every positive integer $n$.

Therefore, the largest value of $k$ is 3.

Remark: It is easily seen that $4 n^{4}+4 n^{2}$ and the next two natural numbers also form a suitable triple for any positive integer $n$, since

$$
\begin{gathered}
4 n^{4}+4 n^{2}=\left(2 n^{2}\right)^{2}+\left(2 n^{2}\right)^{2} \\
4 n^{4}+4 n^{2}+1=\left(2 n^{2}+1\right)^{2}+0^{2} \\
4 n^{4}+4 n^{2}+2=\left(2 n^{2}+1\right)^{2}+1^{2}
\end{gathered}
$$

István Peremiczki (Budapest, Berzsenyi D. Gymnasium, 2nd grade)