## Task $2 / 88$

The smallest two natural numbers $n$ are sought, which are the sum of both two and three consecutive square numbers.
According to the problem statement, we have

$$
n=(m-1)^{2}+m^{2}+(m+1)^{2}=k^{2}+(k+1)^{2}
$$

with $n ; m ; k \in N, n ; m ; k \geq 1$. One can quickly see that $m=k=1$ provides a (though trivial) solution: $n=0^{2}+1^{2}+2^{2}=1^{2}+2^{2}=5$.

Obviously, this is the smallest number with the required property. To find the next largest, we rearrange: $2 k(k+1)=3 m^{2}+1$.

Since $3 m^{2}+1 \equiv 1(\bmod 3)$, it follows that $k \equiv 1(\bmod 3)$, so $k=3 s+1$ with $s \in N$. Substitution and further rearrangement yield

$$
6 s(s+1)+1=m^{2} \quad ; \quad m=\sqrt{6 s(s+1)+1}
$$

If $n$ is to be minimal, then $m ; k ; s$ must also be minimal. We therefore try, starting with the smallest value, the $s$-values to see if they yield natural $m$-values.

For $s=0$, we obtain the already determined (trivial) solution; for $s=1 ; 2 ; 3$, $m$ is not natural, but for $s=4$, we get $m=11, k=13$ and

$$
n=10^{2}+11^{2}+12^{2}=13^{2}+14^{2}=365
$$

Two things are surprising about this result: 1. One obtains five consecutive square numbers (it can be shown that this case is unique).

2. It results in the number of days in a common year.

Additional: The next larger $n$ are $n=108^{2}+109^{2}+110^{2}=133^{2}+134^{2}=35645$ and $n=1078^{2}+$ $1079^{2}+1080^{2}=1321^{2}+1322^{2}=3492725$.