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.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $2$
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $12$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $2$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $12$
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): deficient number