# Are there known solutions to this functional equation involving the square root? $A(x)=c\frac{A\left(\sqrt{x}\right)+A\left(-\sqrt{x}\right)}{2}+x^k$

Consider the following functional equation $$A(x)=c\frac{A\big(\sqrt{x}\big)+A\big(-\sqrt{x}\big)}{2} + x^k,$$ where $$c\in\mathbb R$$ and $$k\in\mathbb N$$ are constants, and $$c$$ is chosen to make $$A(x)$$ analytic (if I am not mistaken, we can find through the root test that the radius of convergence of $$A$$ is fine for $$1). The question is: are there known "nice" solutions? Of course I am looking for a closed formula or solutions in integral form, infinite products, and so on.

Notice that the coefficients $$\{a_n\}_{n\ge 0}$$ in $$A(x)=\sum_{n\ge 0}a_nx^n$$ verify the nice relation $$a_n = c\cdot a_{2n} +\delta(n-k),$$ where $$\delta(m)$$ is equal to $$1$$ if $$m=0$$ and $$0$$ otherwise. Then, writing $$k=o\cdot 2^h$$ with $$o$$ an odd number and $$h\ge 0$$, the only nonzero coefficients are $$a_{o\cdot 2^i}$$ for $$i=0,1,\dots$$ and they are related in a straightforward way, that depends on $$i$$ being less or greater than $$h$$, to $$a_o$$.

I have briefly checked standard references on functional equations without success.

There is an easy to check polynomial solution. For $$k=2^ho$$ as above,
$$A(x):=\sum_{j=0}^h c^{h-j}x^{2^jo}=x^k+cx^{k/2}+c^2x^{k/4}+\dots+c^hx^{k/2^h}$$ Note that the only odd-degree term is the last one, $$c^hx^o$$.
$$\sim *\sim$$ If $$\bf |c|<1$$, this is also the only locally bounded solution on $$\mathbb C$$; more generally, for any $$r\ge1$$ and any bounded $$\mathbb C$$-valued function $$f$$ on the disk $$B(0,r)$$ of radius $$r$$, the linear functional equation $$A(x)=c\frac{A(\sqrt x)+A(-\sqrt x)}{2}+f(x)$$ has a unique bounded solution on $$B(0,r)$$ given inverting the operator $$I-cH$$ with $$(Hg)(x):=\frac{g(\sqrt x)+g(-\sqrt x)}{2}=\frac12\sum_{z^2=x}g(z),$$ the mean value of $$g$$ on the square roots of $$x$$. So the iterates of $$H$$ are given by $$H^n g(x)=\frac1{2^n}\sum_{z^{2^n}=x}g(z),$$ and give the mean value of $$g$$ on the $$2^n$$-th roots of $$x$$. Since $$\|H\|\le 1$$ one has the Neumann expansion $$(I-cH)^{-1}=\sum_{n=0}^\infty H^n$$ and finds $$A(x):=\sum_{n\in\mathbb N, z^{2^n}=x} (c/2)^n f(z).$$ If $$f$$ is a locally bounded function on $$\mathbb C$$, the latter is the only locally bounded solution on $$\mathbb C$$.