Deriving Lorentz transformations I know this is gonna sound backward reasoning. But still, just for the sake of curiosity:
Is it possible to use just Lorentz Contraction and Time Dilation along with the lorentz factor $\gamma$ to derive the Lorentz transformations from one inertial frame to another? If not, what other assumptions are necessary?
 A: We assume that Lorentz transformations are linear and that the origin in one frame maps to the origin in another frame (we do this in the usual derivation of Lorentz transformations as well!). Then, the general form is
$$
t' = a t + b x\,,\qquad x' = c t + d x \,.
$$
Since the transformations are linear, this also implies
$$
\Delta t' = a \Delta t + b \Delta x\,,\qquad \Delta x' = c \Delta t + d \Delta x \,.
$$
Time Dilation
This states that if $\Delta x = 0$, then $\Delta t' = \gamma \Delta t$. Thus, $a=\gamma$. Inversely, if $\Delta x' = 0$ then $\Delta t = \gamma \Delta t'$. This implies $d = \gamma ( ad - b c )$.
Length Contraction
This states that if $\Delta t' = 0$, then $\Delta x' = \frac{ \Delta x}{ \gamma }$. This implies $a=\gamma(ad-bc)$. Inversely, if $\Delta t = 0$, then $\Delta x = \frac{ \Delta x'}{\gamma}$. This implies $d=\gamma$. Using these equations, we can obtain
$$
a = d = \gamma \,,\qquad b c  = \gamma^2 - 1 = \frac{\gamma^2 v^2}{C^2}
$$
EDIT - I'm using $C$ for the speed of light so as not to confuse with the coefficient $c$. 
With this, we can write a generic Lorentz transformation as
$$
t' = \gamma   t - \frac{\gamma v}{C^2 } f(v)  x \,,\qquad x' = \gamma x - \frac{\gamma v}{f(v)} t\, .
$$
for some function $f(v)$. We can now study some properties of this function. 
First, we note the inverse Lorentz transform takes the form
$$
t = \gamma t' + \frac{\gamma v}{C^2} f(v) x' \,,\qquad x= \gamma x' + \frac{ \gamma v }{ f(v) } t' 
$$
The two must be related via $v \to - v$. Thus, we must have $f(v) = f (-v)$.
I don't think we can say more about $f(v)$ without additional input! Let me add one additional input and complete the derivation. 
Constancy of Light Speed
This requires that if $x=Ct$ then $x'=Ct'$. This immediately implies $f(v)=1$. 
