Divisor Sum of Square-Free Integer
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Theorem
Let $n$ be an integer such that $n \ge 2$.
Let $n$ be square-free.
Let the prime decomposition of $n$ be:
- $\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i = p_1 p_2 \cdots p_r$
Let $\map {\sigma_1} n$ be the divisor sum of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.
Then:
- $\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} p_i + 1$
Proof 1
We have that the Divisor Sum Function is Multiplicative.
From the definition of prime number, each of the prime factors of $n$ is coprime to all other divisors of $n$.
From Divisor Sum of Prime Number, we have:
- $\map {\sigma_1} {p_i} = p_i + 1$
Hence the result.
$\blacksquare$
Proof 2
From Divisor Sum of Integer:
- $\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where each of the $k_i$s are equal to $1$;
Hence:
- $\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^2 - 1} {p_i - 1}$
But from Difference of Two Squares:
- $\dfrac {p_i^2 - 1} {p_i - 1} = \dfrac {\paren {p_i + 1} \paren {p_i i 1} } {p_i - 1} = p_i + 1$
Hence the result.
$\blacksquare$
Examples
$\sigma_1$ of $66$
- $\map {\sigma_1} {66} = 144$
$\sigma_1$ of $70$
- $\map {\sigma_1} {70} = 144$
$\sigma_1$ of $110$
- $\map {\sigma_1} {110} = 216$
$\sigma_1$ of $170$
- $\map {\sigma_1} {170} = 324$