Consider the equation $\Delta^2 f=0$ for $f \in C^4(\mathbb R^n)$
$f$ is supposed to be rotationally invariant $f(x)=g(\vert x \vert)$
Prove that all rotation-invariant solutions are of the form $f(x)=c_1+c_2r^2+c_3r^{2-n}+c_4r^{4-n}$ where $r=\vert x\ \vert$ and $n\neq2,4$
What I got:
The solutions to the Poisson equation $\Delta f=0$ under above conditions are:
$$f(x)=c_1r^{2-n}+c_2$$ for any $n\ge2$
How can I get the solutions for $\Delta^2f=0$ from here ?
You have the solutions to $\Delta f = 0$ directly. Those will also be solutions to $\Delta^2 f=0$. What you get that are new for $\Delta^2$ but not $\Delta$ are solving either:
$$ \Delta f = r^{2-n}\\ \Delta f = 1\\ $$
The second gives
$$ \frac{\partial^2 f}{\partial r^2} + \frac{N-1}{r} \frac{\partial f}{\partial r} = 1\\ $$
Plugging in $f=r^2$ gives
$$ 2 + \frac{N-1}{r} 2r = 2N\\ $$
So $\frac{1}{2N} r^2$ will solve the second.
Check that something of the form $c_1 r^{4-N}$ will solve the first.
$$ (4-N)(4-N-1) c_1 r^{4-N-2} + \frac{N-1}{r} (4-N) c_1 r^{4-N-1} = c_1 ((4-N)(4-N-1) + (4-N)(N-1)) r^{2-N} = r^{2-N}\\ c_1 = \frac{1}{((4-N)(4-N-1) + (4-N)(N-1))} $$
So take the two that you already found $r^{2-N}$ and $1$ and the new ones $\frac{1}{2N} r^2$ and $c_1 r^{4-N}$ and any linear combination of them will be a solution to $\Delta^2 f=0$
EDIT:
If $\Delta^2 f = 0$ then $\Delta f$ must be in the kernel of the $\Delta$ operator. You already figured out what the kernel of the $\Delta$ operator was (within rotation invariant functions). So reduced to solving $\Delta f = k$ for k general element of that kernel. Linearity of $\Delta$ ensures this has been done with just the 2 computations we have done.