Belarusian Mathematical Olympiad

a) Let $N$ be a nice number, i.e. $N=d_1^2+d_2^2+d_3^2$, where $d_1,d_2,d_3$ are distinct divisors of $N$. If some divisor of $N$ is divisible by $3$, then $N$ is divisible by $3$.

So we suppose that $d_1,d_2,d_3$ are not divisible by $3$. Then their squares are congruent to $1$ modulo $3$, i.e. $d_1^2=3k_1+1$, $k_1\in \mathbb{N}$, $i=1,2,3$. Therefore $$N=d_1^2+d_2^2+d_3^2=3(k_1+k_2+k_3)+3,$$ and so $N$ is divisible by $3$.

b) There exists a nice number $N^{\prime }$. For example, if $N=30$ and its divisors are $d_1=1$, $d_2=2$, $d_3=5$, then $$d_1^2+d_2^2+d_3^2=1^2+2^2+5^2=1+4+25=30=N,$$ i.e. $N$ is nice.

Consider $N(p)=N{p}^2$, where $p$ is some positive integer and $p>1$. If $d_1,d_2,d_3$ are distinct divisors of $N$, then it is obvious that $d_{1}p,d_{2}p,d_{3}p$ are distinct divisors of $N(p)$, and $$N(p)=N{p}^2=[N=d_1^2+d_2^2+d_3^2]=(d_1^2+d_2^2+d_3^2)p^2=(d_{1}p{)}^2+(d_{2}p{)}^2+(d_{3}p{)}^2.$$ From this equality it follows that $N(p)$ is nice too. Since any positive integer can be used as $p$, there are infinitely many nice numbers $N(p)$.