Help understanding the solution of $m^2 = n^3 -4$. The material I'm studying has the solution for $m^2 = n^3 + 4$ but I can't quite understand it. It starts like this:
First suppose, for a contradiction, that $m$ is odd. Then $n^3 = m^2 − 4 = (m + 2)(m − 2)$. Any common factors of $m + 2$ and $m − 2$ divide $(m + 2) − (m − 2) = 4$ and so,  $\gcd(m + 2, m − 2) = $1$.
It follows that $m + 2$ and $m − 2$ are both cubes.
a) How did they get to the conclusion $\gcd(m+2, m-2)=1$? I mean, at this stage of the proof, I don't know.
$m\neq 10$ so I could think if $m=10$ then gcd(12,8) $=4\neq 1$.
b) Why does it follow that $m+2$ and $m-2$ are both cubes, and not that their multiplication is a cube?
Let's carry on the proof.
[...] No two odd cubes differ by 4, giving a contradiction.
Since $m$ is even there exists $m'$ such that $m = 2m'$. We thus have that $4m'^2 = n^3 + 4$ and $n$ is divisible by $2$ so we write $n = 2n'$. Now, $m'^2 = 2n'^3 + 1$ and so $m'$ is odd so let $m' = 2r + 1$. Substituting this into the previous equation tells us that $2r^2 + 2r = n'^3$.
No problems with the above, let's carry on
Finally,$r^2 + r = r (r + 1)$ and since $r$ and $r + 1$ are coprime we must have that one of $r$ and $r + 1$ is a cube and the other is four times a cube.
c) Why are we considering $r^2+r$ when we have $2r^2+2r=n'^3$? Yes, they have a common factor of $2$, but we don't know this about $n'$ do we? we can't cancel them.
d) ``since $r$ and $r + 1$ are coprime we must have that one of $r$ and $r + 1$ is a cube and the other is four times a cube'' ????
Any help with the above is really appreciated.

a) You do know $m \neq 10$ because you have assumed $m$ odd. From "any common factors of $m+2$ and $m-2$ divide $(m+2)-(m-2)=4$," we can conclude that the only possible common factors of $m+2$ and $m-2$ are $1$, $2$, or $4$. But if $m$ is odd, then so are $m+2$ and $m-2$. So $2$ and $4$ cannot be divisors. This leaves $1$ as the greatest common divisor.
b) If $a$ and $b$ are not themselves cubes, yet $ab=c^3$ is a cube, then $a$ and $b$ must share a common factor greater than $1$. An example (not a proof) would be $3\cdot 9 = 27=2^3$, and $3$ and $9$ share a common factor of $3$. To see why this holds, consider the prime factorization of $c$ and note that each prime must appear with an exponent that is a multiple of $3$. If $a$ and $b$ are not cubes, then they each have some prime divisor that appears with an exponent not a multiple of $3$. Then you can show $a$ and $b$ must share a prime divisor.
c) Nothing has been canceled, just factoring $n'^3$ as $2r(r+1)$.
d) See the answer to b). The reasoning is similar. We know that $2$ divides $n'^3$, so $8$ must divide $n'^3$. Only one of $r$ and $r+1$ is even, so $4$ must divide one of them. So if we divide $2r(r+1)=n'^3$ by $8$, we get $\frac{r(r+1)}{4}=n''^3$ where $n'=2n''$. This is a product of two coprime integers (either $\frac{r}{4}$ and $r+1$ or $r$ and $\frac{r+1}{4}$) equal to a cube, hence both factors are cubes. Thus one of $r$ and $r+1$ is a cube and the other is $4$ times a cube.
