I keep searching but I can't find any place that gives a good method of showing a process is NOT Markov. The definition I am using is that for every $s<t$ and $g$ bounded borel there is $f$ borel such that $E[g(X_t)|\mathcal{F}_s] = f(X_s)$ a.s.
I was told that $(B_t+1)^2$ is not a Markov process relative to the filtration generated by $B_t$ (standard Brownian Motion), I would like to show this.
Negating the definition, I need to find $s<t$ and $g$ bounded borel such that there is no function $f$ with $E[g((B_t+1)^2)|\mathcal{F}_s] = f((B_s+1)^2)$ a.s.
By the Markov property for $B_t$ i see that I can write $$ E[g((B_t+1)^2)|\mathcal{F}_s] = h(B_s) $$ where $$ h(x) = E[g((B_t-B_s+x+1)^2)] $$ but I don't see how I can show that for a particular $g$ this can't be rewritten as $f((B_s+1)^2)$.
A bounded Borel function $h$ on $\mathbb{R}$ is a function of $(x+1)^2$ if and only if $h(x)=g(x+1)$ for some symmetric function $g$, i.e., $g(x)=g(-x)$.
Let $f$ be any bounded Borel function and define the symmetric function $g(x)=f(x^2)$ and let $h(x)=g(x+1)=f((x+1)^2)$. Since the transition function $p_t$ for Brownian motion is symmetric, we have for convolution $(p_t*g)(x)=(p_t*g)(-x)$ and since it is also shift invariant, we get $(p_t*h)(x)=(p_t*g)(x+1)$ so $(p_t*h)(x)=\phi_t((x+1)^2)$ for some bounded Borel $\phi_t$.
Let $(B_t)$ be standard Brownian motion with filtration $({\cal F}_t)$, and define $X_t:=(B_t+1)^2$. For $s<t$ we have $$\mathbb{E}\left[f(X_t)\,\big|\,{\cal F}_s\right] = \mathbb{E}\left[h(B_t)\,\big|\,{\cal F}_s\right] =(p_{t-s}*h)(B_s)=\phi_{t-s}(X_s).$$
So in fact, $(X_t)$ is a Markov process.