If we have a function $f(x)$, for which we know that $f(p^k)=(p^s+1)^k p^{sk}$ where $p$ is prime, $k$ is a real number, and $s$ is a constant, how do we find $f(x)$? My try: let $k=\log_p(x)$, so that no matter what $p$ is, $p^k$ is constant equal to $x$. Plugging this in to the other side of the equation, we have $$f(p^{\log_p(x)})=(p^s+1)^{\log_p(x)}p^{s\log_p(x)}$$ $$=x^{\log_p(p^s+1)}x^s$$ $$=x^{\log_p(p^s+1)+s}$$. The only problem with this is that this side of the equation is not constant if $p$ changes, e.g. $x^{\log_2(2^s+1)+s}\not=x^{\log_3(3^s+1)+s}$, while on the other side it was a constant expression (see above). This seems to be a problem, and I can't find the error.
Suggestions?
Consider a "simpler" version of your functional equation where $p$ can be any sort of number
In that case:
$$f(p^k) = {\lbrace( (p^s + 1)p^s \rbrace)}^k$$
That reduces to:
$$f(p^k) = f(p)^k$$
So now we have :
$$f(p) = (p^s + 1)p^s $$
Note here that if we consider $f(p^k)$it then must be true that:
$$(p^{sk}+ 1)p^{sk} = ((p^s + 1)p^s)^k $$
This is clearly not true for all $p,k$ for any given $s$ and therefore we immediately conclude the real number version of your functional equation has no global solution for all real numbers.
This argument can now be modified to consider prime numbers and real exponents to again show the original problem has no solution.