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 $n \neq 2$ ?


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 $n=2$. 
