# Three times Number whose Sigma is Square/Proof 1

## 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

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.

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

$\blacksquare$