How to show that for $n \in \mathbb{N}$, $n^2 + 4n - 8$ is square if and only if $n=2$ Let $n$ be a positive integer which is bigger than $1$.

How to show that $n^2 + 4n - 8$ is not square if $n \neq 2$ ?

 A: Assume there is such $n$ that $n>1$ and $n \neq 2$, yet $n^2+4n-8$ is a square, which implies $n>2$. 
Then $$(n+2)^2=n^2+4n+4>n^2+4n-8>n^2$$
Since $n>2$. So $n^2+4n-8$ is a square number stuck between $(n+2)^2$ and $n^2$, which means it can only be $(n+1)^2$. So we have$$n^2+4n-8 =(n+1)^2=n^2+2n+1 \implies n=\frac{9}{2}$$
This is a contradiction, since $n \in \mathbb{N}$. So we can only have $n=2$. 
A: Suppose $n^2 + 4n - 8=m^2\,$, then for $n^2 + 4n - 8-m^2 = 0\,$ to have an integer root, its reduced discriminant needs to be a perfect square. But $\Delta'=4+8+m^2=m^2+12\,$, and for $m \gt 5\,$ it can be easily verifed that $m^2 \lt m^2+12 \lt (m+1)^2\,$ i.e. $\,m^2+12\,$ lies strictly between two consecutive squares, so it cannot be a perfect square itself. This leaves the cases $m \le 5$ to be checked by hand, where the only perfect square is found to be for $m=2$ with $2^2+12=16=4^2\,$, which gives $n=-2 \pm \sqrt{\Delta'}=-2 \pm 4\,$. Since $n \in \mathbb{N}$ the only eligible root is $n=2\,$.
A: Let $n^2+4n-8=k^2$. Obviously $n,k$ are both odd or even.
$$(k-n)(k+n)=4(n-2)$$
Let $k-n=2u$, $k+n=2v$
$$4uv=4(v-u-2)\iff uv-v+u+2=0\iff(u-1)(v+1)=-3$$
Checking factors of $-3$, we get $(u,v)=(0,2)\implies(n,k)=(2,2)$ as the only solution.
