Hyperbolas: Deriving $\frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1$ from $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$ My textbook's section on Hyperbolas states the following:

If the foci are $F_1(-c, 0)$ and $F_2(c, 0)$ and the constant difference is $2a$, then a point $(x, y)$ lies on the hyperbola if and only if $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$.
To simplify this equation, we move the second radical to the right-hand side, square, isolate the remaining radical, and square again, obtaining $\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2 - c^2} = 1$.

I've attempted to "move the second radical to the right-hand side, square, isolate the remaining radical, and square again", but I cannot derive $\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2 - c^2} = 1$. I could be misunderstanding the instructions, but my attempts to derive the textbook's solution by precisely following the instructions have not been successful.
I would greatly appreciate it if people could please take the time to demonstrate the derivation of $\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2 - c^2} = 1$ from $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$, as mentioned in the textbook. Are the textbook's instructions incorrect/insufficient or am I simply misunderstanding them?
 A: You have\begin{multline*}\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}=\pm2a\Longleftrightarrow\\\Longleftrightarrow(x+c)^2+y^2+(x-c)^2+y^2-2\sqrt{(x+c)^2+y^2}\sqrt{(x-c)^2+y^2}=4a^2.\end{multline*}This is the same thing as saying that$$\sqrt{(x+c)^2+y^2}\sqrt{(x-c)^2+y^2}=-2a^2+c^2+x^2+y^2.$$Squaring both sides, one gets$$\bigl((x+c)^2+y^2\bigr)\bigl((x-c)^2+y^2\bigr)=(-2a^2+c^2+x^2+y^2)^2$$This is equivalent to$$-a^4+c^2 a^2+x^2 a^2+y^2 a^2-c^2 x^2=0$$or$$(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)\text,$$and this means that$$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1.$$Of course, a little extra work is required in order to prove that the two expressions are equivalent.
A: $$\sqrt{(x+c)^2 + y^2} - \sqrt{(x-c)^2 + y^2} = 2a$$
$$\Rightarrow \left(\sqrt{(x+c)^2 + y^2} - \sqrt{(x-c)^2 + y^2}\right)^2 = 4a^2$$
$$\Rightarrow (x+c)^2 + y^2 + (x-c^2) + y^2 - 4a^2 = 2\sqrt{\left((x+c)^2+y^2\right)\left((x-c)^2 + y^2\right)}$$
$$\Rightarrow 2x^2 + 2c^2 + 2y^2 - 4a^2 = 2\sqrt{\left((x+c)^2+y^2\right)\left((x-c)^2 + y^2\right)}$$
$$\Rightarrow x^2 + c^2 + y^2 - 2a^2 = \sqrt{\left((x+c)^2+y^2\right)\left((x-c)^2 + y^2\right)}$$
$$\Rightarrow (x^2 + c^2 + y^2 - 2a^2)^2 = \left((x+c)^2+y^2\right)\left((x-c)^2 + y^2\right)$$
$$\Rightarrow (x^2 + c^2 + y^2 - 2a^2)^2 = ((x^2 + c^2 + y^2) + 2xc)((x^2 + c^2 + y^2)- 2xc)$$
In the RHS, we can apply the formula: $(a-b)(a+b) = a^2 - b^2$
So, we get:
$$(x^2 + c^2 + y^2 - 2a^2)^2 = (x^2 + y^2 + c^2)^2 - 4x^2c^2$$
$$\Rightarrow (x^2+c^2 + y^2)^2 + 4a^4 - 4a^2(x^2+c^2 + y^2) = (x^2 + y^2 + c^2)^2 - 4x^2c^2$$
$$\Rightarrow  a^4 - a^2(x^2+c^2 + y^2) = - x^2c^2$$
$$\Rightarrow  x^2(a^2-c^2) + a^2y^2 =   a^2(a^2 - c^2)$$
$$\Rightarrow  \frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} =   1$$
To show that the two expressions are equivalent, you must show that the implications where you squared both side are reversible.
A: Let's do the case
$$
\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = 2a
$$
which holds provided $x>0$.
Move the second radical to the right-hand side and square:
$$
(x + c)^2 + y^2=4a^2+(x - c)^2 + y^2 + 4a\sqrt{(x - c)^2 + y^2}
$$
Lots of terms cancel:
$$
2cx=4a^2-2cx+4a\sqrt{(x - c)^2 + y^2}
$$
which becomes
$$
cx-a^2=a\sqrt{(x - c)^2 + y^2}
$$
Note that $cx\ge a^2$, so we can square again and arrive easily to the result.
The case $x<0$ is identical.
