B1. Find all natural numbers $n$ whose cube is equal to the sum of the squares of three not necessarily distinct divisors of the number $n$.
B1. Let $a, b$ and $c$ be divisors of the number $n$, for which $n^{3}=a^{2}+b^{2}+c^{2}$. Then $a, b, c \leq n$, so $n^{3}=a^{2}+b^{2}+c^{2} \leq 3 n^{2}$ or $n \leq 3$. The case $n=1$ is excluded, as 1 is not the sum of three natural numbers. If $n=2$, the only divisors are 1 and 2. However, none of the sums $1^{2}+1^{2}+1^{2}, 1^{2}+1^{2}+2^{2}$, $1^{2}+2^{2}+2^{2}$ and $2^{2}+2^{2}+2^{2}$ equals $2^{3}$, so the case $n=2$ is also excluded. The only solution is $n=3$, since $3^{3}=3^{2}+3^{2}+3^{2}$.

2. Let $a \leq b \leq c$ be divisors of the number $n$, for which $n^{3}=a^{2}+b^{2}+c^{2}$. Since $c^{2}$ divides $n^{3}$, it must also divide $a^{2}+b^{2}$. However, by our assumption, $a^{2}+b^{2} \leq 2 c^{2}$, so we have two possibilities, $a^{2}+b^{2}=c^{2}$ or $a^{2}+b^{2}=2 c^{2}$. In the first case, $n^{3}=2 c^{2}$ or $\left(\frac{n}{c}\right)^{2} n=2$, which means that $n$ divides 2. In the second case, $n^{3}=3 c^{2}$ or $\left(\frac{n}{c}\right)^{2} n=3$, which means that $n$ divides 3. It follows that $n$ is equal to 1, 2, or 3. In the same way as in the first solution, we show that the solution is only $n=3$.[^0]