Three times Number whose Sigma is Square/Proof 1

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Let $n \in \Z_{>0}$ be a positive integer.

Let the $\sigma$ value of $n$ be square.

Let $3$ not be a divisor of $n$.

Then the $\sigma$ value of $3 n$ is square.


Let $\sigma \left({n}\right) = k^2$.

We have from Numbers whose $\sigma$ is Square:

$\map \sigma 3 = 4 = 2^2$

As $3$ is not a divisor of $n$, it follows that $3$ and $n$ are coprime.


\(\displaystyle \sigma \left({3 n}\right)\) \(=\) \(\displaystyle \sigma \left({3 n}\right) \sigma \left({3 n}\right)\) Sigma Function is Multiplicative
\(\displaystyle \) \(=\) \(\displaystyle 2^2 k^2\) from above
\(\displaystyle \) \(=\) \(\displaystyle \left({2 k}\right)^2\) from above