Divisor Sum of Non-Square Semiprime/Examples/115

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Example of Divisor Sum of Non-Square Semiprime

$\map {\sigma_1} {115} = 144$

where $\sigma_1$ denotes the divisor sum function.


Proof 1

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 $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.


We have that:

$115 = 5 \times 23$

Hence:

\(\ds \map {\sigma_1} {115}\) \(=\) \(\ds \frac {5^2 - 1} {5 - 1} \times \frac {23^2 - 1} {23 - 1}\)
\(\ds \) \(=\) \(\ds \frac {4 \times 6} 4 \times \frac {22 \times 24} {22}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds 6 \times 24\)
\(\ds \) \(=\) \(\ds \paren {2 \times 3} \times \paren {2^3 \times 3}\)
\(\ds \) \(=\) \(\ds 2^4 \times 3^2\)
\(\ds \) \(=\) \(\ds \paren {2^2 \times 3}^2\)
\(\ds \) \(=\) \(\ds 12^2\)
\(\ds \) \(=\) \(\ds 144\)

$\blacksquare$


Proof 2

We have that:

$115 = 5 \times 23$

and so by definition is a semiprime whose prime factors are distinct.


Hence:

\(\ds \map {\sigma_1} {115}\) \(=\) \(\ds \paren {5 + 1} \paren {23 + 1}\) Divisor Sum of Non-Square Semiprime
\(\ds \) \(=\) \(\ds 6 \times 24\)
\(\ds \) \(=\) \(\ds \paren {2 \times 3} \times \paren {2^3 \times 3}\)
\(\ds \) \(=\) \(\ds 2^4 \times 3^2\)
\(\ds \) \(=\) \(\ds \paren {2^2 \times 3}^2\)
\(\ds \) \(=\) \(\ds 12^2\)
\(\ds \) \(=\) \(\ds 144\)

$\blacksquare$