Product of Coprime Numbers whose Sigma is Square has Square Sigma

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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

$\blacksquare$