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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 $N$?

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    The above question is posted as a Challenge problem on Brilliant.org, which offers weekly problem sets to test student's problem solving abilities. John Chang has been posting questions on math.stackexchange.com and expecting others to solve the problems for him. He has posted another one of our questions at http://math.stackexchange.com/questions/266337/smallest-possible-value-on-fibonacci-function - Calvin Lin Mathematics Challenge Master2012-12-29

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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:

  • $N = 1 + 2^2 + 4^2 + 8^2$ - 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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$N$ 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 $N=130$.