How can $n^5+4$ be a perfect square? How can one find all $n \in \mathbb{N}$ such that  $n^5+4$ is a perfect square?
I see that $n^5=(x+2)(x-2)$ here im suck can someone help ?

Partial solution:
$d=\gcd(x+2,x-2)$ is $1$, $2$ or $4$.
If $d=1$ then $x+2$ and $x-2$ are perfect fifth powers. That is clearly impossible.
If $d=2$ then $x+2=2a$ and $x-2=2^4b=16b$ where $a$ is odd and $b$ is an integer (or vice versa). Either way, we have that $x=a+8b$, which is odd; a contradiction.
If $d=4$ then $x+2=4a$ and $x-2=4a-4$, where $a$ is some integer, that is, $x=4a-2$. Then $n$ is even, that is, $n=2m$ and
$$2m^5=a^2-a$$
There is at least one example, for $a=2$.
