Prove this is a Martingale Prove that 
$$Z_t:=\frac{e^{W_t^2/(1+2t)}}{\sqrt{1+2t}}$$
 is a $\mathscr{F}_t$-martingale.
I have tried all the usual manipulations without any success.
The only useful fact I think should be used is that:
if $X_t\sim N(\mu,\sigma^2)$ then:
$\mathbb{E}[e^{\alpha\frac{X^2_t}{\sigma^2}}]=\frac{e^{(\frac{\mu^2\alpha}{\sigma^2(1-2\alpha)})}}{\sqrt{1-2\alpha}}(1)$
EDIT: I will answer my question.
 A: Hint: Since $W_t=W_s+\sqrt{t-s}\,Y$ where $Y$ is standard normal and independent of $\mathscr F_s$, $$E(Z_t\mid\mathscr F_s)=\frac{g(W_s,\sqrt{t-s},1+2t)}{\sqrt{1+2t}},\qquad g(w,a,b)=E(\mathrm e^{(w+aY)^2)/b}).$$ Hence the task is to compute the function $g$ and even, more precisely, to show that, for every $(w,a,b)$ such that $b\gt2a^2$, $$g(w,a,b)=\frac{\mathrm e^{w^2/(b-2a^2)}}{\sqrt{b-2a^2}}.$$
A: I don't know much about probabilities and its well-known formulas so let me try it from scratch.
Assume $Z\sim N(\mu,\sigma^2)$ and let us compute $E\left[e^{Z^2}\right]$.
$$\begin{align}E\left[e^{Z^2}\right]&=\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}e^{x^2}e^{-\frac{\left(x-\mu\right)}{2\sigma^2}}dx\\&=\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}e^{-\frac{\left(x-\frac{\mu}{1-2\sigma^2}\right)^2-\frac{2\sigma^2}{(1-2\sigma^2)}}{2\sigma^2/(1-2\sigma^2)}}dx\\&=\frac{e^{\frac{\mu^2}{1-2\sigma^2}}}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}e^{-\frac{y^2}{2\sigma^2/(1-2\sigma^2)}}dy\\&=\frac{e^{\frac{\mu^2}{1-2\sigma^2}}}{\sqrt{1-2\sigma^2}}\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}e^{-s^2/2\sigma^2}\\&=\frac{e^{\frac{\mu^2}{1-2\sigma^2}}}{\sqrt{1-2\sigma^2}}\end{align}$$
Summarizing

$$\begin{equation}Z\sim N(\mu,\sigma^2)\implies E\left[e^{Z^2}\right]=\frac{e^{\frac{\mu^2}{1-2\sigma^2}}}{\sqrt{1-2\sigma^2}}\end{equation}\ \ \ \ \ \ \ (1)$$


We try now to compute $E\left[Z_t|F_s\right]$. Let's see
$$\begin{align}E\left[Z_t|F_s\right]&=E\left[\frac{e^{W_t^2/(1+2t)}}{\sqrt{1+2t}}| F_s\right]\\&=\frac{1}{\sqrt{1+2t}}E\left[e^{\left(W_t-W_s+W_s\right)^2/(1+2t)}| F_s\right]\\&=\frac{e^{W_s^2/(1+2t)}}{\sqrt{1+2t}}E\left[e^{(W_t-W_s)^2/(1+2t)}e^{2(W_t-W_s)W_s/(1+2t)}|F_s\right]\end{align}$$
To shorten notation let me put $Z:=(W_t-W_s)/\sqrt{1+2t}$ and $X_s:=W_s/\sqrt{1+2t}$. With this notation we continue the computation of the conditional expectation above.
$$\begin{align}E\left[E^{Z^2}E^{2ZX}|F_s\right]&=E\left[\sum_{k=0}^{\infty}\frac{2^kX_2^kZ^k}{k!}e^{Z^2}|F_s\right]\\&=\sum_{k=0}^{\infty}E\left[X_s^k\frac{2^kZ^k}{k!}e^{Z^2}|F_s\right]\\&=\sum_{k=0}^{\infty}X_s^k(\omega')E\left[\frac{2^kZ^k(\omega)}{k!}e^{Z^2}|F_s\right]\\&=\sum_{k=0}^{\infty}X_s^k(\omega')E_\omega\left[\frac{2^kZ^k(\omega)}{k!}e^{Z^2}\right]\\&=\sum_{k=0}^{\infty}E_\omega\left[\frac{2^kX_s^k(\omega')Z^k(\omega)}{k!}e^{Z^2}\right]\\&=E_\omega\left[e^{Z^2(\omega)}e^{2X_s(\omega')Z(\omega)}\right]\end{align}$$
(I wrote explicitly in the few last lines the state variables and a subscript on the expectation to indicate with respect to which this is computed).
Putting this in the previous computation we get
$$E\left[Z_t|F_s\right]=\frac{1}{\sqrt{1+2t}}E_\omega\left[e^{(Z+X_s(\omega'))^2}\right]$$
Now we apply the formula (1) to $Z+X_s(\omega')\sim N\left(\frac{W_s(\omega')}{\sqrt{1+2t}},\frac{t-s}{1+2t}\right)$ to finally get
$$E\left[Z_t|F_s\right]=\frac{1}{\sqrt{1+2t}}e^{W_s^2/(1+2s)}\frac{\sqrt{1+2t}}{\sqrt{1+2s}}=Z_s.$$
A: $\mathbb{E}[Z_t|\mathscr{F_s}]= \mathbb{E}[exp{(W_t-W_s+W_s)^2/(1+2t)}|\mathscr{F_s}]/\sqrt{1+2t}$.
Now call $Y:=W_t-W_s$. One has that $Y\sim N(0,t-s)$ and $Y$ is independent of $\mathscr{F_s}$. Moreover since $W_s$ is $\mathscr{F_s}$-measurable, we can treat it like a constant and therefore one has:
$\tilde Y := W_s + Y \sim N(W_s, t-s)$. $\tilde Y$ is therefore independent of $\mathscr{F_s}$ and we get:
$\mathbb{E}[Z_t] = \mathbb{E}[e^{\tilde Y^2/(1+2t)}]/\sqrt{1+2t}= \mathbb{E}[e^{\tilde Y^2\cdot\frac{t-s}{1+2t}\frac{1}{t-s}}]/\sqrt{1+2t} $.
Now we can apply (1) with $\alpha = \frac{t-s}{1+2t}$ yielding the desired result, hence:
$\mathbb{E}[Z_t|\mathscr{F_s}]= Z_s$.
