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: a) "At this stage of the proof I don't know $m\ne 10$" - Actually, at this stage you know (or temporarily assume) that $m$ is odd
b) Claim. If $ab=c^3$ with $\gcd(a,b)=1$, then $a,b$ are both cubes. Proof: Asssume otherwise and let $(a,b,c)$ be a counterexample with minimal $c$. The certainly $c>1$ as there cannot be a counterexample with $c=1$. Then there exists a prime $p\mid c$. Then $p\mid ab$, so $p\mid a$ or $p\mid b$. Wlog. $p\mid a$. But then $p\nmid b$ (coprime!). Then from $p^3\mid c^3$ we find that in fact $p^3\mid a$. Then $(\frac a{p^3},b,\frac cp)$ is a smaller counterexample, contradiction. $\square$
c),d): From  $2r(r+1)=n^3$ note that one of $r,r+1$ is odd, so either $n^3=(2r)\cdot(r+1)$ or $n^3=r\cdot(2r+2)$ is a factorization of a cube into two coprime factors. As above, each factor must be a cube. And if $2x=y^3$, then $y$ is even so $y=2z$ and finally $x=4\cdot z^3$ is four times a cube
A: 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.
A: 
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.
