Three times Number whose Divisor Sum is Square/Proof 1
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Theorem
Let $n \in \Z_{>0}$ be a positive integer.
Let the divisor sum $\map {\sigma_1} n$ of $n$ be square.
Let $3$ not be a divisor of $n$.
Then the divisor sum of $3 n$ is square.
Proof
Let $\map {\sigma_1} n = k^2$.
We have from $\sigma_1$ of $3$:
-
- $\map {\sigma_1} 3 = 4 = 2^2$
As $3$ is not a divisor of $n$, it follows that $3$ and $n$ are coprime.
Thus:
\(\ds \map {\sigma_1} {3 n}\) | \(=\) | \(\ds \map {\sigma_1} {3 n} \map {\sigma_1} {3 n}\) | Divisor Sum Function is Multiplicative | |||||||||||
\(\ds \) | \(=\) | \(\ds 2^2 k^2\) | from above | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2 k}^2\) | from above |
$\blacksquare$