Three times Number whose Divisor Sum is Square/Proof 2

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

From Numbers whose Divisor Sum is Square:

$\map {\sigma_1} 3 = 4 = 2^2$

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

$\blacksquare$