# 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

- 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $24$