<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] The equation $$N = 1 + d_1^2 + d_2^2 + d_3^2$$ implies $N$ even by reducing $\mod 2$. Suppose $2^2|N$ then there are three possible cases:


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*$N = 1 + 2^2 + 4^2 + 8^2$ - but this is impossible by arithmetic.

*$N = 1 + 2^2 + 4^2 + p^2$ (with $p>3$) - but reducing $\mod p$ we find $0 \equiv 13 \pmod p$ so $p = 13$ but this too is impossible by arithmetic.

*$N = 1 + 2^2 + 3^2 + 4^2$ - impossible by arithmetic.


So $2$ is the highest power of $2$ dividing $N$, thus we have the cases:


*

*$N = 1 + 2^2 + p^2 + q^2$ (with $q < 2p$) but this is impossible by reducing $\mod 2$

*$N = 1 + 2^2 + p^2 + (2p)^2$ so $N = 5(1+p^2)$ and $5|N$ so $p$ must be $3$ (impossible by arithmetic) or $5$.


This shows the only possible solution is $130$.
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