Three times Number whose Sigma is Square

From ProofWiki
Jump to navigation Jump to search

Theorem

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.


Proof 1

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

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

$\sigma \left({3}\right) = 4 = 2^2$


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

Thus:

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

$\blacksquare$


Proof 2

From Numbers whose $\sigma$ is Square:

$\sigma \left({3}\right) = 4 = 2^2$


The result follows as a specific instance of Product of Coprime Numbers whose Sigma is Square has Square Sigma.

$\blacksquare$