Smallest Pair of Quasiamicable Numbers

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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.

$\blacksquare$


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