Is it true that spring has more force acting on it at its positive maximum amplitude than than at the negative one? Am I missing something?

It seems obvious to me that at $+A$ and $-A$, the spring has restorative forces equal in magnitude but opposite in direction. 
But since gravity is always pulling it down, the spring at position $-A$ must have less net force acting on it. But my book says that at both positions, the spring has its maximum $\sum F = ma$. How does this make sense?
 A: The equilibrium position in this case is not where the spring is not stretched, it is actually stretched by a $\Delta x$ amount with $F_{spring}(0) = k\Delta x$. 
So the spring force on point A is a little smaller than in point -A, since $ F_{spring}(A) = -k(A-\Delta x)$ and  $ F_{spring}(-A) = k(A+\Delta x)$ so it compensates the "extra" force.
You have to notice that in this equilibrium position 
$F_{spring} - mg = 0$ , 
so
$F_{spring} = k\Delta x = mg$
with  
$\Delta x = mg/k$.
Substituting in  
$ F_{net}(A) = F_{spring}(A) - mg = -k(A-\Delta x) - mg = -k(A - \frac{mg}{k}) - mg = -kA $
the same hold for the -A position
$ F_{net}(-A) = F_{spring}(-A) - mg = -k(-A-\Delta x) - mg = -k(-A-\frac{mg}{k}) - mg = kA $
A: In accordance with Hooke's law the force is linear with distance. Incorporating gravity only means that the equillibirum position of the spring has changed, the "zero" around which it oscillates. The gravitational pull is already compensated by the spring. Thus the magnitude of the force is euqal at $-A$ and $+A$.
Edit: When the gravitational pull on the mass on the spring is considered, the spring elongates. This results in a new equillibirum position $x'_0 = x_0 + \Delta x = x_0 + \frac{m g}{k}$. Since the force is always (in Hooke's regieme) linear with distance, you can just neglect the gravitational force since it is compensated by the spring. It is a simple superposition of forces.
A: The answer is no. The net force at the maximum elongation points has the same magnitude.
This is because the rest point of the spring is modified by the gravitational weight of the mass. The mass oscillates around this new rest point, and at the points of maximum amplitude the net force is the same. One can tell that the gravitational force
can be "integrated out".
To see this in more details, consider the net force on the mass, which in this case is the sum of the elastic force and the gravitational force (which is constant):
$$
F=mg- k (x-x_0)
$$
where $x_0$ is the rest position of the spring (without the mass).
Now you can calculate the rest position $x_0'$ of the spring including the effect of gravitation, which is defined as the point where $F=0$. So you obtain that at the position $x=x_0'$ you have
$$
0=mg- k (x_0'-x_0)\Rightarrow x_0'=x_0+mg/k
$$
Now, what happens if one changes coordinates?
$$
F=mg- k (x-x_0)=mg- k [x-(x_0'-mg/k)]=mg- k [x-x_0']-mg=-k(x-x_0')
$$
which means that, if one considers the displacement with respect to the new rest point $x_0'$, the force is simply given by
$$
F=-k(x-x_0')
$$
Now it is easy to see that at the two points of maximum elongation the net force is the same in magnitude, but opposite in sign.
Edit:
Consider the maximum elongation $A$, which is measured with respect to the new rest point $x_0'$. One can reverse the transformation I'v done and obtain that the net force at $x-x'_0=\pm A$ (i.e., $x=x'_0\pm A$) is 
$$
F=mg- k (x'_0\pm A-x_0)=\mp k(x-x_0')
$$
Note that the net force is the same in magnitude, but the elongation of the spring with respect to the original rest point $x_0$ (without considering the mass $m$) is not the same but it is $(x'_0-x_0\pm A)$.
