Smallest Pair of Quasiamicable Numbers
Jump to navigation
Jump to search
Theorem
The smallest pair of quasiamicable numbers is $48$ and $75$.
Proof
From Quasiamicable Numbers: $48$ and $75$ we have that $48$ and $75$ are quasiamicable numbers.
It remains to be demonstrated that these are indeed the smallest such.
Let $n$ be the smaller number of the quasiamicable pair.
Then we must have $\map {\sigma_1} n - n - 1 > n$.
Since $\map {\sigma_1} n > 2 n$, $n$ is abundant.
The abundant numbers for $n < 48$ are:
- $12, 18, 20, 24, 30, 36, 40, 42$
And we have:
\(\ds \map {\sigma_1} {12} - 13\) | \(=\) | \(\ds \paren {1 + 2 + 4} \paren {1 + 3} - 13\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 15\) | ||||||||||||
\(\ds \map {\sigma_1} {15} - 16\) | \(=\) | \(\ds \paren {1 + 3} \paren {1 + 5} - 16\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 8\) | \(\ds \ne 12\) | |||||||||||
\(\ds \map {\sigma_1} {18} - 19\) | \(=\) | \(\ds \paren {1 + 2} \paren {1 + 3 + 9} - 19\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 20\) | ||||||||||||
\(\ds \map {\sigma_1} {20} - 21\) | \(=\) | \(\ds \paren {1 + 2 + 4} \paren {1 + 5} - 21\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 21\) | \(\ds \ne 18\) | |||||||||||
\(\ds \map {\sigma_1} {21} - 22\) | \(=\) | \(\ds \paren {1 + 3} \paren {1 + 7} - 22\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 10\) | \(\ds \ne 20\) | |||||||||||
\(\ds \map {\sigma_1} {24} - 25\) | \(=\) | \(\ds \paren {1 + 2 + 4 + 8} \paren {1 + 3} - 25\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 35\) | ||||||||||||
\(\ds \map {\sigma_1} {35} - 36\) | \(=\) | \(\ds \paren {1 + 5} \paren {1 + 7} - 36\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 12\) | \(\ds \ne 24\) | |||||||||||
\(\ds \map {\sigma_1} {30} - 31\) | \(=\) | \(\ds \paren {1 + 2} \paren {1 + 3} \paren {1 + 5} - 31\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 41\) | ||||||||||||
\(\ds \map {\sigma_1} {41} - 42\) | \(=\) | \(\ds 0\) | \(\ds \ne 30, 36\) | |||||||||||
\(\ds \map {\sigma_1} {36} - 37\) | \(=\) | \(\ds \paren {1 + 2 + 4} \paren {1 + 3 + 9} - 37\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 41\) | ||||||||||||
\(\ds \map {\sigma_1} {40} - 41\) | \(=\) | \(\ds \paren {1 + 2 + 4 + 8} \paren {1 + 5} - 41\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 49\) | ||||||||||||
\(\ds \map {\sigma_1} {49} - 50\) | \(=\) | \(\ds \paren {1 + 7 + 49} - 50\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 7\) | \(\ds \ne 40\) | |||||||||||
\(\ds \map {\sigma_1} {42} - 43\) | \(=\) | \(\ds \paren {1 + 2} \paren {1 + 3} \paren {1 + 7} - 43\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 53\) | ||||||||||||
\(\ds \map {\sigma_1} {53} - 54\) | \(=\) | \(\ds 0\) | \(\ds \ne 42\) |
hence $n$ is at least $48$, proving the result.
$\blacksquare$
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $48$