I'm trying to figure out where my logic is wrong here. Probably I have a dumb mistake somewhere, but I cannot find it on my own.
Let $X_t$ be a stochastic process adapted to the filtration $\mathcal{F}_t$. We say $X_t$ is a martingale if $\mathbb{E}[X_t \mid \mathcal{F}_s] = X_s$ for all $s < t$.
I'm looking to prove that the process $X_t = B_t^2$, where $B_t$ is standard Brownian motion, is not a martingale. I happen to know that this is indeed the case, because it is well known that $B_t^2 - t$ is a martingale, and further by Itô's lemma one can show that the drift term does not vanish, which is a necessary condition for $X_t$ to be a martingale here.
However, I'm able to produce the following (incorrect) proof as follows. \begin{equation} \mathbb{E}[X_t | \mathcal{F}_s] = X_s \iff \mathbb{E}[X_t - X_s | \mathcal{F}_s] = 0 \\ \end{equation} And \begin{align} \mathbb{E}[X_t - X_s | \mathcal{F}_s] &= \mathbb{E}[B_t^2 - B_s^2 | \mathcal{F}_s] \\ & = \mathbb{E}[(B_t - B_s)(B_t+B_s) | \mathcal{F}_s] \\ & = \mathbb{E}[B_t - B_s]\mathbb{E}[B_t + B_s | \mathcal{F}_s] \text{ (by independent increments)} \\ & = 0 \text{ as } B_t - B_s \sim \mathcal{N}(0, t-s) \end{align} Hence we get $\mathbb{E}[X_t | \mathcal{F}_s] = X_s$.
I assume there is some subtlety to using independent increments, but surely $B_t - B_s$ is independent of $\mathcal{F}_s$? Seems to be part of the definition of Brownian motion or a Wiener process.
Thanks