If I have a functional equation defined in $f : N → N$ and I have to show that there are no solutions. If I show it for $f : R → R$, does it directly imply there is no solution in $f : N → N$?
Question part 2: even if I would substitute 0 for some variable and the result would show there are no solutions in $f : R → R$, does that mean there are no solutions in $f : N → N$? Seems odd to me
For example, the "functional equation" $f(x)^2 = \cos(2\pi x)$ has solutions for functions $\mathbb N \to \mathbb N$, but not for $\mathbb R \to \mathbb R$.