As far as I have grasped the concept,
$$ y[n] = \left( 2 x[n] - x^2[n] \right)^2 $$
is a memoryless system because even if we give negative values of $n$, we still get the overall result in the positive sign, right?
And the following expression is memorable system because we get the negative answer for negative values of $n$, right?
$$y[n] = \left( 2 x[n] - x^2[n] \right)$$
Confusion may arise, for causal systems, from mistaking "having negative signal values (amplitudes)" and "depending on negative time indices".
A strict memory-less system depends neither on past (for the causal case) nor on future values of the input but on current ones to determine the current value of the output. The output at $n$ only depends on inputs at $n$. For a strict memory-less and causal system, you could not depend on former (negative, in a way relative to the current index) time indices.
By extension, some allow memory-less systems to have a constant delay, i.e., to hold only one value at a given lag: the output at $n$ only depends on inputs at $n-k$ with a fixed integer $k$. So, in general, a memory-less system can have the following form, with $k$ any integer:
$$y[n] = f(x[n-k])\,.$$
For realizable causal systems that cannot look ahead in the future, only $k \ge 0$ are allowed.
The memoryless system means a system which does not use past values of the input. There is nothing to do with the sign of your output. That just means there isn't any storage element in your system.
Although reverse is not necessarily true every time, that implies also casuality.
