Divisor of Deficient Number is Deficient
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Theorem
Let $n$ be a perfect number.
Let $n = k d$ where $r$ is a positive integer.
Then $k$ is deficient.
Proof
We have by definition of divisor sum function and perfect number that:
- $\dfrac {\map {\sigma_1} {k d} } {k d} < 2$
But from Abundancy Index of Product is greater than Abundancy Index of Proper Factors:
- $\dfrac {\map {\sigma_1} {k d} } {k d} > \dfrac {\map {\sigma_1} k} k$
That is:
- $\dfrac {\map {\sigma_1} k} k < 2$
Hence the result by definition of deficient.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $12$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $12$