<s>[INST] Determining the number $N$ Let $1 = d_1 < d_2 <\cdots< d_k = N$ be all the divisors of $N$ arranged in increasing order. Given that $N=d_1^2+d_2^2+d_3^2+d_4^2$, determine $N$. The divisors include $N$. It seems that $130$ is an answer. Is there another possible answer for $N$?
 [/INST] $N$ is even (if not then all $d_i$ are odd, making $\sum_{i=1}^4 d_i^2$ even).  Therefore $d_1=1$ and $d_2=2$, and at exactly one of $d_3$ and $d_4$ is even.
Suppose that $4 \mid n$.  Then one of $d_3, d_4$ is $4$ and the other is an odd prime $p$.  Since $N=21+p^2$ and $p \mid N$, we have $p \mid 21$.  But $4 \nmid 21+3^2$ and $5 \mid 21+7^2$, ruling out both choices of $p$.
Thus $4 \nmid n$.  $d_3$ is an odd prime $p$ and $d_4$ is even.  Since $d_4/2$ is a smaller divisor, it could only be $d_3$.  Therefore $N = 1+4+p^2+4p^2 = 5(1+p^2)$.  $3$ cannot divide a number of this form, and clearly $5$ does.  Therefore $d_3=5$ and $d_4=10$, uniquely determining $N=130$.
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