Three times Number whose Divisor Sum is Square/Proof 1

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

$\blacksquare$