We denote the sum of divisors function with $$\sigma(m)=\sum_{d\mid m}d.$$
If we suppose that there exist perfect numbers $n$ and $N$ satisfying $\gcd(n,N)=1$, then it is easy to deduce the next claim by direct proof (or deducing a more simple equation similar using the hypotesis and deducing next, that is more elaborated, from the more simple one).
Claim. Suppose that there exist perfect numbers $n$ and $N$ that are coprime, then $$(\sigma(Nn))^2=2n^2(\sigma(N))^2+2N^2(\sigma(n))^2.\tag{1}$$
I add as curiosity, and it is obvious, you can think in $(1)$ as the Pythagorean theorem for right triangles with hypotenuse $\frac{1}{\sqrt{2}}\sigma(Nn)$, and integral sides $n\sigma(N)$ and $N\sigma(n)$.
I don't know if $(1)$ was in the literature (as was said is it possible to write similar and more simple equations invoking our hypothesis; or more complicated using more artificious hypothesis for tuples of perfect numbers). After this motivation I would like to know what can be deduced of the equation $(1)$ for integers $1<a<b$ (I believe for the case of perfect numbers, that I can not deduce more that if $a$ and $b$ are coprime integers satisfying $(1)$, and satisfying that one of our numbers is perfect, then the other integer also is perfect).
Questions.
A) Let $a,b$ two positive integers that satisfy $1<a<b$, the condition $\gcd(a,b)=1$ and $$(\sigma(ab))^2=2a^2(\sigma(b))^2+2b^2(\sigma(a))^2.$$ What can be deduced from here? If you can, please add some simple example of a solution $(a,b)$ of our equation.
B) Let $a,b$ two positive integers that satisfy $1<a<b$, the condition $\gcd(a,b)>1$ and $$(\sigma(ab))^2=2a^2(\sigma(b))^2+2b^2(\sigma(a))^2.$$ What can be deduced from here? If you can, please add some simple example of a solution $(a,b)$ of our equation. Many thanks.
Thus I am asking for general feedback about our equation, about what easy deductions can be done from the hypothesis. After there are some answer for A) and B) I should accept an answer (if it is hard do a deduction for some A) or B) tell us).