I am trying to show that $(X(t)-\lambda t)^2 - \lambda t$ is a martingale, where $X(t)$ is a Poisson process with rate $\lambda$. So far, what I have done is: \begin{align*} E\left((X(t)-\lambda t)^2 - \lambda t| \mathcal{F}_s\right)&= E\left(X(t)^2-2\lambda t X(t) + \lambda^2 t^2 - \lambda t| \mathcal{F}_s\right) \\ &=E\left((X(t)-X(s))^2| \mathcal{F}_s\right) + E\left(2X(t)X(s)| \mathcal{F}_s\right) - E\left(X(s)^2| \mathcal{F}_s\right) - E\left(2X(t)\lambda t| \mathcal{F}_s\right) \\ & + E\left(\lambda^2 t^2| \mathcal{F}_s\right) -\lambda t\\ &=E\left((X(t)-X(s))^2\right) + 2X(s)E\left(X(t)| \mathcal{F}_s\right) - X(s)^2 -\lambda t - 2 \lambda t E\left(X(t)| \mathcal{F}_s\right)+ \lambda^2 t^2\\ &= 2\lambda t - 2\lambda tX(s) + 2X(s)\left(X(s)+ \lambda(t-s)\right) -\lambda t - 2\lambda t\left(X(s) + \lambda (t-s)\right)+ \lambda^2 t^2\\ \end{align*}
I am not sure why my result is not working out. Does anyone have any ideas what I am doign wrong? Thanks!
I think everything you're doing is correct, but perhaps it would benefit from exploiting some cancellations so things work out more smoothly.
In your second line, with six terms, I claim if you take the first, fourth, fifth, and sixth terms and sum them, you get $$\boxed{ \lambda^2 s^2 -\lambda s -2\lambda t X_s}.$$ I used this grouping because lots of things cancel. I also used the fact if $Y$ is a Poisson random variable with mean (parameter) $\mu$, $E(Y^2)=\mu^2+\mu$.
This leaves us with evaluating the conditional expectation of $$2X_sX_t - X_s=X_s(2X_t-X_s)=X_s(2(X_t-X_s) + X_s)=2X_s(X_t-X_s)+X_s^2$$ Conditioning, and noting that the conditional expectation of $X_t-X_s$ is $\lambda(t-s)$ yields $$\boxed{2X_s\lambda(t-s)+X_s^2}.$$
Summing the boxed expressions gives $$(X_s-\lambda s)^2 - \lambda s,$$ which shows the martingale property holds.