Sum of Squares of Divisors of 24 and 26 are Equal

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Theorem

The sum of the squares of the divisors of $24$ equals the sum of the squares of the divisors of $26$:

$\map {\sigma_2} {24} = \map {\sigma_2} {26}$

where $\sigma_\alpha$ denotes the divisor function.


Proof

The divisors of $24$ are:

$1, 2, 3, 4, 6, 8, 12, 24$


The divisors of $26$ are:

$1, 2, 13, 26$


Then we have:

\(\displaystyle \) \(\) \(\displaystyle 1^2 + 2^2 + 3^2 + 4^2 + 6^2 + 8^2 + 12^2 + 24^2\)
\(\displaystyle \) \(=\) \(\displaystyle 1 + 4 + 9 + 16 + 36 + 64 + 144 + 576\)
\(\displaystyle \) \(=\) \(\displaystyle 850\)


\(\displaystyle \) \(\) \(\displaystyle 1^2 + 2^2 + 13^2 + 26^2\)
\(\displaystyle \) \(=\) \(\displaystyle 1 + 4 + 169 + 676\)
\(\displaystyle \) \(=\) \(\displaystyle 850\)

$\blacksquare$


Sources