Product of Coprime Numbers whose Divisor Sum is Square has Square Divisor Sum

Theorem

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

Let the divisor sum of $m$ and $n$ both be square.

Let $m$ and $n$ be coprime.

Then the divisor sum of $m n$ is square.

Proof

Let $\map {\sigma_1} m = k^2$.

Let $\map {\sigma_1} n = l^2$.

Thus:

\(\ds \map {\sigma_1} {m n}\)

\(=\)

\(\ds \map {\sigma_1} m \map {\sigma_1} n\)

Divisor Sum Function is Multiplicative

\(\ds \)

\(=\)

\(\ds k^2 l^2\)

from above

\(\ds \)

\(=\)

\(\ds \paren {k l}^2\)

from above

$\blacksquare$