I was inspired by this question Does a non-trivial solution exist for $f'(x)=f(f(x))$?
And tried coming up with similar problems, one interesting case I found was $f'(x) +f(x)=f(f(x))$ which has a linear solution $f(x) = x+1$
This makes sense because if $f(x)$ is a polynomial then $f(f(x))$ has order $n^2$. Either $f'(x)$ which has order $n-1$ or $f(x)$ which has order $n$ must match that (and $n-1$ cannot for real $n$), implying that $n^2=n$ or $n=1$
Is that enough to solve the problem as if it was a linear differential equation? Or are there other weak solutions that are possible? (maybe ones involving polynomials with complex exponents?)
