Power of Prime is Deficient

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Theorem

Let $n \in \Z_{>0}$ be a power of a prime number $p$:

$n = p^k$

for some $k \in \Z_{>0}$.


Then $n$ is deficient.


Proof

From Divisor Sum of Power of Prime:

$\map {\sigma_1} n = \dfrac {p^{k + 1} - 1} {p - 1}$

Thus:

\(\ds \map A n\) \(=\) \(\ds \map {\sigma_1} n - 2 n\) Definition of Abundance
\(\ds \) \(=\) \(\ds \dfrac {p^{k + 1} - 1} {p - 1} - 2 p^k\)
\(\ds \) \(=\) \(\ds \dfrac {p^{k + 1} - 1 - \paren {2 p^{k + 1} - 2 p^k} } {p - 1}\)
\(\ds \) \(=\) \(\ds \dfrac {2 p^k - p^{k + 1} - 1} {p - 1}\)
\(\ds \) \(=\) \(\ds \dfrac {p^k \paren {2 - p} - 1} {p - 1}\)
\(\ds \) \(=\) \(\ds \dfrac {p^k - 1} {p - 1} - p^k\)
\(\ds \) \(=\) \(\ds \dfrac {p^k} {p - 1} - p^k - \dfrac 1 {p - 1}\)

If $p = 2$ then $\dfrac {p^k} {p - 1} - p^k = 0$ and so:

$\dfrac {p^k} {p - 1} - p^k - \dfrac 1 {p - 1} = - \dfrac 1 {p - 1} < 0$


If $p > 2$ then $\dfrac {p^k} {p - 1} < p^k$ and so:

$\dfrac {p^k} {p - 1} - p^k - \dfrac 1 {p - 1} < 0$

Thus in all cases:

$\map A n < 0$

and so $n = p^k$ is deficient by definition.

$\blacksquare$


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