<s>[INST] 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.
 [/INST] 
a) How did they get to the conclusion gcd(m+2, m-2)=1?$.

Let $\gcd(m+2, m-2) = d$ then $d|m+2, m-2$ so $d|(m+2) - (m-2) = 4$.  So $d = 1,2,4$.  But $m$ is odd (so your example of $m=10$ won't work) so $m\pm 2$ is odd.  So $d \ne 2,4$.  So $d = 1$.
Note: the general idea that if $\gcd(a,b)$ divides $a,b$ it will divide any sum or difference of $a,b$ leads to the Lemma

Lemma:  $\gcd(a,b) = \gcd(a\pm b, b)$

and so $\gcd(m+2, m-2) = \gcd((m+2)-(m-2), m-2) = \gcd(4, m-2)$ which must be a divisor of $4$.

b) Why does it follow that m+2 and m−2 are both cubes, and not that their multiplication is a cube?

Because $m+2$ and $m-2$ have no factors in common.
Any prime factor $p$ of $m+2$ or of $m-2$ will divide into $n^3$ a power of a multiple of $3$.  But as that prime factor belongs to $m+2$ or to $m-2$ alone and not to the other  it must divide into $m+2$ or $m-2$ a power of a multiple of $3$ times.
So $m+2$ and $m-2$ are cubes.

c) Why are we considering r2+r when we have 2r2+2r=n′3?

$2r^2 + 2r = n'^3$ so
$r^2 + r = \frac {n'^3}2$
$r(r+1) = \frac {n^3}2$.
That's why we are considering 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'' ????

$2|n'^3$ so $2|n'$.  And so $8|n'^3$.
Let $n' = 2\overline n$. so $n'^3 = 8\overline n^3$
So $r(r+1) = \frac {8\overline n^3}2 = 4\overline n^3$.
Now one of $r, r+1$ is odd and the other is even.  And the even one is divisible by $4$.  Let $r_1 = $ the odd one, and let $4r_2 =$ the even one we have $4r_1r_2 =4\overline{n}^3$ or $r_1r_2 = \overline n^3$.
Now we just redo what we did in b).  $r, r+1$ are relatively prime so $r_1, 4r_2$ are relatively prime and $r_1, r_2$ are relatively prime.  And if we redo the argument of b) we know that $r_1, r_2$ are both perfect cubes.
So $r_1$ and $4r_2$, which are $r, r+1$ (although we do not know which is which) are so that $r_1$ is a perfect cube and $4r_2$ is $4$ times a perfect cube.
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