Functional Square Root

[Catullus]

What about that (x) = f(sqrt(x))? Is f(x) = x^4 the only answer to it?

Tommy is correct in his deduction, as far as I can tell. No reason to doubt him. But \(\infty\) is also a solution if you ask for meromorphic solutions. So there are two constant solutions \(0\) and \(\infty\), but other than that, the only other solution is \(x^4\), and Tommy's proof is good enough.

Hmm, I saw some weird formatting on your reply.