Sum of Squares of Divisors of 24 and 26 are Equal

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: