Timeline for What is the solution, $f(n)$, of the following functional equation: $mf(m)+nf(n)=(m+n+xmn)f(m+n+xmn)$?

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Dec 20, 2017 at 1:29	history	edited	Somos	CC BY-SA 3.0	Added note about definition of h().
Dec 20, 2017 at 0:26	comment	added	LSpice		What is $h$ in your second paragraph? It looks like you're defining $g$, but the rest of your exposition suggests you still want $g : n \mapsto n f(n)$. Are you defining $h$ by the equation $g(n) = h(\phi(n))$? If so, then is the idea that $h$ is defined arbitrarily on numbers that aren't $1$ modulo $x$? (In the same vein, what does $g((n - 1)/x)$ mean when $n \not\equiv 1 \pmod x$?)
Dec 19, 2017 at 4:45	vote	accept	mark
Dec 19, 2017 at 0:54	history	edited	Somos	CC BY-SA 3.0	Added phi(n)^k k>0 proviso.
Dec 18, 2017 at 4:54	vote	accept	mark
Dec 19, 2017 at 4:45
Dec 17, 2017 at 18:26	history	edited	Somos	CC BY-SA 3.0	Added sentence about \|f(t)\|>=b.
Dec 16, 2017 at 17:58	history	edited	Somos	CC BY-SA 3.0	Fixed my typo.
Dec 16, 2017 at 17:53	history	answered	Somos	CC BY-SA 3.0