Given a vector field $V=\xi\partial_x+\eta\partial_u$, I computed the second prolongation of $V$ to be $$Pr^{(2)}V=V+\left[D_x\eta-u_xD_x\xi\right]\partial_{u_x}+\left[D_xD_x\eta-2u_{xx}D_x\xi-u_xD_xD_x\xi\right]\partial_{u_x}$$ where $D_x=\partial_x+u_x\partial_u+u_{xx}\partial_{u_x}+\cdots.$ Now I am asked to show that the equation $$u_{xx}=0$$ admits an 8-dimensional group of Lie point symmetries. I don't know what this would look like mathematically - I know a Lie point symmetry $\psi^s:(x,u)\to\psi^s(x,u)=(\tilde x,\tilde u)$ is one such that $$\tilde u_{\tilde x\tilde x}=0\iff u_{xx}=0,$$ but what does it mean by an 8-dimensional group? And what does this look like mathematically?
Edit 1:
I recalled that $\psi^s$ is a Lie Point symmetry iff $Pr^{(2)}V (u_{xx})=0$ whenever $u_{xx}=0$. This leads to $$\eta_{xx}+2u_x\eta_{xu}+{u_x}^2\eta_{uu}=u_x(\xi_{xx}+2u_x\xi_{xu}+{u_x}^2\xi_{uu})$$ I don't know what to do next.
Edit 2:
I was experimenting with $\eta(x,u),\xi(x,u)$ which satisfy the above 2nd order PDE, and was able to find 8 different ones, namely:
$\eta=1$
$\xi=1$
$\eta=x$
$\xi=x$
$\eta=u$
$\xi=u$
$\eta=ux,\xi=x^2$
$\xi=ux,\eta=u^2$.
How would I find these without just guessing? How can I show that there are no more?
You can rearrange your equation to get $$\eta_{xx}+(2\eta_{xu}-\xi_{xx})u_x+(\eta_{uu}-2\xi_{xu})(u_x)^2-\xi_{uu}(u_x)^3=0$$ From $u_{xx}=0$ we should have $u_x$ to be an arbitrary constant. Thus above equation should hold for every $u_x$. So we would have $4$ p.d.e.'s \begin{align} \eta_{xx}&=0&(1)\\ 2\eta_{xu}-\xi_{xx}&=0&(2)\\ \eta_{uu}-2\xi_{xu}&=0&(3)\\ \xi_{uu}&=0&(4) \end{align} From $(1)$ & $(4)$ we get $$\eta=f_1(u)x+g_1(u)\ ,\ \ \ \xi=f_2(x)u+g_2(x)$$ Substitute these into $(2)$ & $(3)$: \begin{align} 2f_1'(u)&=f_2''(x)u+g_2''(x)&(5)\\ 2f_2'(x)&=f_1''(u)x+g_1''(u)&(6) \end{align} LHS of $(5)$ has no $x$ dependency, so $f_2''$ and $g_2''$ can only be some constants. Similarly for $f_1''$ and $g_1''$ in $(6)$. So we have $f_i$, $g_i$ in the form of $$ax^2+bx+c$$ Substitute these back into $(5)$ and $(6)$, I got \begin{align} \eta&=(a_1u+a_2)x+a_5u^2+a_3u+a_4\\ \xi&=(a_5x+a_6)u+a_1x^2+a_7x+a_8 \end{align} Which has dimension $8$.