Product of Coprime Numbers whose Sigma is Square has Square Sigma

Theorem

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

Let the $\sigma$ value of $m$ and $n$ both be square.

Let $m$ and $n$ be coprime.

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

Proof

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

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

Thus:

\(\displaystyle \sigma \left({m n}\right)\)

\(=\)

\(\displaystyle \sigma \left({m}\right) \sigma \left({n}\right)\)

Sigma Function is Multiplicative

\(\displaystyle \)

\(=\)

\(\displaystyle k^2 l^2\)

from above

\(\displaystyle \)

\(=\)

\(\displaystyle \left({k l}\right)^2\)

from above

$\blacksquare$