Sum of squares of divisors

Number Theory Level 4

Let $n$ be a natural number.

Suppose $d_{1},d_{2},d_{3}, \ldots$ are the divisors of $n$ such that

$1=d_{1}<d_{2}<d_{3}<\ldots$

Suppose $n=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}$

Find the value of $n$ .

1 solution

Souryajit Roy
May 8, 2014

First show that $n$ is even,but not divisible by $4$

Then its clear $d_{2}=2$ and $d_{3}$ is the least possible prime divisor of $n$ .Let $d_{3}=p$

Then $d_{4}=2p$

Hence, $n=5(p^{2}+1)$ .

The only possibility is $p=5$ and so $n=1^{2}+2^{2}+5^{2}+10^{2}=130$

one cannot assume that d_(4) is equal to 2p. The answer takes invalid assumption

Nisshith Sharma - 6 years, 1 month ago

The problem previously stated $1 = d_1 < d_2 < d_4$ . I have since corrected it to $1 = d_1 < d_2 < d_3$ .

Calvin Lin Staff - 6 years, 5 months ago

This doesn't cover the cases where $p<d_4<2p$ .

Jared Low - 6 years, 5 months ago