We call the number n a "special number" if there are three distinct natural numbers divisors (of n) so that the sum of their squares is equal to n. We know that n is a natural number and n is diffrent from 0. 1. Demonstrate that any "special number" is divisible by 3. 2. Show that there is an infinity of natural "special numbers".
P.S. Sorry for writing mistakes, I'm a beginner.
So $n=a^2+b^2+c^2$ with $a\mid n$, $b\mid n$, $c\mid n$.
If any of $a,b,c$ is a multiple of $3$, then so clearly is $n$. In all other cases, $a^2\equiv b^2\equiv c^2\equiv 1\pmod 3$ so that $n\equiv 1+1+1\equiv 0\pmod 3$.
Start with any three distinct naturals $a_0,b_0,c_0$ and let $n_0=a_0^2+b_0^2+c_0^2$. In general, $a_0\nmid n_0$. Let $\frac{n_0}{a_0}=\frac uv$ in shortest terms and assume that the denominator $v$ is $>1$. Let $a_1=va_0$, $b_1=vb_0$, $c_1=vc_0$, $n_1=v^2n_0$. Then $a_1,b_1,c_1$ are three distinct naturals and $n_1=a_1^2+b_1^2+c_1^2$, but now $a_1\mid b_1$. Also note that the denominator of $\frac{n_1}{b_1}$ is not bigger than that of $\frac{n_0}{b_0}$ etc. Thus we can repeat the above for $b$ and for $c$, arriving ultimately at a special number.
More concretely, $126=3^2+6^2+9^2$ is special and so is $126k^2=(3k)^2+(6k)^2+(9k)^2$ for all $k$.