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:
\(\ds \) | \(\) | \(\ds 1^2 + 2^2 + 3^2 + 4^2 + 6^2 + 8^2 + 12^2 + 24^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + 4 + 9 + 16 + 36 + 64 + 144 + 576\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 850\) |
\(\ds \) | \(\) | \(\ds 1^2 + 2^2 + 13^2 + 26^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + 4 + 169 + 676\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 850\) |
$\blacksquare$
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $24$