Suppose $m^3=n^4-4$ where $m,n \in \mathbb Z$.
a) Show that $m$ cannot be even if $n$ is odd.
b) Show that $m$ and $n$ cannot both be even.
c) By considering the prime factors of $m$,$n^2-2$ and $n^2+2$, show there is no integer solutions to $m^3=n^4-4$
I can do the first two parts, but find it hard to start part c, how shall I think about this problem and where shall I start?
On
a) if n is odd : n^2 = 1 (mod 8) , n^4 = 1 (mod 8)
n^4 - 4 = 1 - 4 = -3 (mod 8) while m^3 = 0 (mod 8) if m is even
b) if m and n are even : m^3 = 0(mod 8) , n^4 - 4 = -4 (mod 8)
so they can't be equal
On
$m^3=n^4-4$
first part
$$n=2k+1 \to m^3=n^4-4=(2k+1)^4-4=odd-even=odd \to m^3=odd \to m=odd=2q'+1 $$
second part
$$m=2q,n=2k \to \\ m^3=n^4-4 \to (2q)^3=(2k)^4-4 \to 8q^3=16k^4-4\\8q^3=4(4k^2-1) $$ divide by 4 $$2q=4k^2-1 \\even=odd $$ that's impossible
3rd part
by 2nd part m,n can't be even
by 1st part m,n are odd may have solution
so
if n=odd $(n^2+2,n^2-2)=1 \to $ so write $m^3=m*m^2=m.m.m=(n^2+2)(n^2-2)$
but $(m,m^2) \neq1 ,(m,m,m) \neq 1 $ this fact show $(n^2+2,n^2-2) \neq 1$
so equation has not integer solution
From a) and b), $m$ must be odd and hence so must be $n^4-4=(n^2+2)(n^2-2)$. As $\gcd(n^2+2,n^2-2)\mid 4$, this implies that $n^2\pm2$ are coprime, hence each of them must be a cube. Hence we have two cubes that differ by $4$.