Set of all $n$; $n={d^2_1 + d^2_2 + d^2_3 +d^2_4}$

Question

$A$ is the set of all $n$ numbers where $n={d^2_1 + d^2_2 + d^2_3 +d^2_4}$. Here $1=d_1<d_2<d_3<d_4$ where $d_1,d_2,d_3,d_4$ are the $4$ smallest divisors of $n$. As an example $130=1^2+2^2+5^2+10^2$ belongs to $A$. What we can say about following statements?

1. If $n$ belongs to $A$, then $n$ is even.
2. $A$ has a $n$ value which is divisible by $4$.
3. $130$ is the one and only value which is divisible by $5$, that belongs to $A$.

Addition (from the comments): I found this question in an old challenge exam. The translation is my own and may need polishing.

Answer 1

We can show that $n=130$ is the only integer with this property.

Let $p<q<r$ be the three smallest prime divisors of $n$ (if it has less prime divisors then the latter primes simply won't exist).

1. It is not possible that $n$ is odd. For then all four smallest divisors of $n$ are also odd. But then the sum of their squares is divisible by four, implying that $n$ should be even after all. So $p=2$.
2. It is not possible that $2$ is the only prime divisor of $n$. For then its four smallest divisors would be $1,2,4$ and $8$, but the sum of their squares is odd.
3. It is not possible that the four smallest divisors of $n$ are $1,2,4$ and $q$. If this were the case we would have $$ n=1+4+16+q^2\equiv2\pmod 4, $$ so $n$ is not divisible by four. A contradiction.
4. If the four smallest divisors of $n$ are $1,2,q$ and $2q$, then $$ n=1+4+q^2+4q^2=5+5q^2. $$ For $n$ to be divisible by $q$ this implies that $q\mid 5$. So we must have $q=5$ and $n=130.$
5. If the four smallest divisors of $n$ are $1,2,q,r$ then $$ n=1+4+q^2+r^2 $$ is an odd integer contradicting the fact that $p=2\mid n$.

The above list covers all the possibilities for the four smallest divisors, so we are done.