Sum of Large and Small divisors of 24n + 1

 Suppose $N = 24n + 1$ is a positive integer, which is not a perfect square. ($n$ is also a positive integer).

A large divisor of $N$ is a divisor of $N$ which is $ > \sqrt{N}$.

A small divisor of $N$ is a divisor of $N$ which is $ < \sqrt{N}$.

Let $L$ be the sum of the large divisors of $N$ and let $S$ be the sum of the small divisors of $N$.

Show that $$ L = S  \mod 24$$ 

For eg, $145 = 29 \times 5$ has the divisors $1, 5, 29, 145$ and $145 + 29 = 6 = 1 + 5 \mod 24$.

Scroll down for a solution.

.

.

.

.

Solution:

Any number $a$ which is relatively prime to $24$ is of the form $a = 6m \pm 1$.

Now $$a^2 - 1 = 36m^2 \pm 12m = 12m (3m \pm 1) = 0 \mod 24$$

This is because $m(3m \pm 1)$ is an even number (either $m$ is even, or $3m \pm 1$ is).

Now since $N$ is not a perfect square, we can pair off the divisors of $N$ as $sl = N$ with $s$ being a small divisor and $l$ being a large divisor.

Note that both $s$ and $l$ are relative prime to $24$, with $sl = 1 \mod 24$.

Now $s^2 - sl =  1 - 1 = 0 \mod 24$, i.e $s(s-l) = 0 \mod 24$. Since $\gcd(s, 24) = 0$, we must have that $s = l \mod 24$.

Thus $\sum s = \sum l \mod 24$.

$\textbf{Additonal Remarks:}$ Similarly we can show that the sum of divisors of a number of the form $24 n - 1$ is divisible by $24$.